C++博客-syhd142-文章分类-ACM/ICPC

HDU 3065 病毒侵袭持续中

Fucker — Mon, 10 Jan 2011 07:58:00 GMT

模板题

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <queue>

#define KIND 26
#define N 1005
#define M 2000005

struct node
{
    node *fail;
    node *next[KIND];
    int id;
    node ()
        {
            fail = NULL;
            id = 0;
            memset(next, 0, sizeof(next));
        }
};

char ch[M], key[N][55];
int count[N];

void insert(node *&root, char *ch, int num)
{
    node *p = root;
    int i = 0, t;
    while(ch[i])
    {
        t = ch[i] - 'A';
        if(!p->next[t]) p->next[t] = new node();
        p = p->next[t];
        i++;
    }
    p->id = num;
}

void AC(node *&root)
{
    std::queue<node *> q;
    q.push(root);
    while(!q.empty())
    {
        node *p = NULL;
        node *t = q.front();
        q.pop();
        for(int i = 0; i < KIND; i++)
        {
            if(t->next[i])
            {
                p = t->fail;
                while(p)
                {
                    if(p->next[i])
                    {
                        t->next[i]->fail = p->next[i];
                        break;
                    }
                    p = p->fail;
                }
                if(!p) t->next[i]->fail = root;
                q.push(t->next[i]);
           }
        }
    }
}

void query(node *&root, char *ch)
{
    int i = 0, t;
    node *p = root, *tmp;
    while(ch[i])
    {
        if(ch[i] > 'Z' || ch[i] < 'A')
        {
            p = root;
            i++;
            continue;
        }
        t = ch[i] - 'A';
        while(!p->next[t] && p != root) p = p->fail;
        p = p->next[t];
        if(!p) p = root;
        tmp = p;
        while(tmp != root)
        {
            count[tmp->id]++;
            tmp = tmp->fail;
        }
        i++;
    }
}

int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        node *root = new node();
        memset(count, 0, sizeof(count));
        for(int i = 1; i <= n; i++)
        {
            scanf("%s", key[i]);
            insert(root, key[i], i);
        }
        AC(root);
        scanf("%s", ch);
        query(root, ch);
        for(int i = 1; i <= n; i++)
        {
            if(count[i])
            {
                printf("%s: %d\n", key[i], count[i]);
            }
        }
    }
    return 0;
}

Fucker 2011-01-10 15:58 发表评论

HDU 2896 病毒侵袭

Fucker — Sun, 09 Jan 2011 16:11:00 GMT

模板题

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

#define KIND 128
#define M 10010

struct node
{
    node *fail;
    node *next[KIND];
    int id;
    node ()
        {
            fail = NULL;
            id = 0;
            memset(next, 0, sizeof(next));
        }
};

char ch[M];
queue<node *> q;
vector<int> g;

void insert(node *&root, char *ch, int id)
{
    node *p = root;
    int i = 0, t;
    while(ch[i])
    {
        t = ch[i] - 31;
        if(!p->next[t]) p->next[t] = new node();
        p = p->next[t];
        i++;
    }
    p->id = id;
}

void AC(node *&root)
{
    q.push(root);
    while(!q.empty())
    {
        node *p = NULL;
        node *t = q.front();
        q.pop();
        for(int i = 0; i < KIND; i++)
        {
            if(t->next[i])
            {
                p = t->fail;
                while(p)
                {
                    if(p->next[i])
                    {
                        t->next[i]->fail = p->next[i];
                        break;
                    }
                    p = p->fail;
                }
                if(!p) t->next[i]->fail = root;
                q.push(t->next[i]);
           }
        }
    }
}

bool query(node *&root, char *ch, int count)
{
    g.clear();
    int i = 0, t, top = 0, flag = 0;
    node *p = root, *tmp;
    while(ch[i])
    {
        t = ch[i] - 31;
        while(!p->next[t] && p != root) p = p->fail;
        p = p->next[t];
        if(!p) p = root;
        tmp = p;
        while(tmp != root && tmp->id)
        {
            flag = 1;
            g.push_back(tmp->id);
            tmp = tmp->fail;
        }
        i++;
    }
    if(!flag) return false;
    sort(g.begin(), g.end());
    printf("web %d:", count);
    printf(" %d", g[0]);
    for(int i = 1; i < g.size(); i++)
    {
        if(g[i] != g[i - 1])
        printf(" %d", g[i]);
    }
    printf("\n");
    return true;
}

int main()
{
    int n, total;
    while(~scanf("%d", &n))
    {
        node *root = new node();
        total = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%s", ch);
            insert(root, ch, i + 1);
        }
        AC(root);
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", ch);
            if(query(root, ch, i + 1)) total++;
        }
        printf("total: %d\n", total);
    }
    return 0;
}

Fucker 2011-01-10 00:11 发表评论

HDU 2222 Keywords Search

Fucker — Wed, 05 Jan 2011 09:38:00 GMT

由于论文的需要，看了一下AC自动机的原理，感觉就是trie树+kmp，这两个单独都写过，

但是合在一起还真没写过，参照网上的代码写了一个HDU上的题目，水的很，入门级别的。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 500005
#define KIND 26

struct node
{
    node *fail;
    node *next[KIND];
    int count;
    node()
        {
            fail = NULL;
            count = 0;
            memset(next, 0, sizeof(next));
        }
}*queue[N];

void insert(node *root, char *str)
{
    int i = 0, index;
    node *p = root;
    while(str[i])
    {
        index = str[i] - 'a';
        if(p->next[index] == NULL) p->next[index] = new node();
        p = p->next[index];
        i++;
    }
    p->count++;
}

void build(node *root)
{
    int front = 0, rear = 0;
    root->fail = NULL;
    queue[rear++] = root;
    while(front < rear)
    {
        node *tmp = queue[front++];
        node *p = NULL;
        for(int i = 0; i < KIND; i++)
        {
            if(tmp->next[i])
            {
                p = tmp->fail;
                while(p)
                {
                    if(p->next[i])
                    {
                        tmp->next[i]->fail = p->next[i];
                        break;
                    }
                    p = p->fail;
                }
                if(!p) tmp->next[i]->fail = root;
                queue[rear++] = tmp->next[i];
            }
        }
    }
}

char str[1000005];

int query(node *root)
{
    int i = 0, ans = 0, index;
    node *p = root, *tmp;
    while(str[i])
    {
        index = str[i] - 'a';
        while(p->next[index] == NULL && p != root) p = p->fail;
        p = p->next[index];
        if(!p) p = root;
        tmp = p;
        while(tmp != root)
        {
            ans += tmp->count;
            tmp->count = 0;
            tmp = tmp->fail;
        }
        i++;
    }
    return ans;
}

int main()
{
    int t, nkey, ans;
    char key[100];
    scanf("%d", &t);
    while(t--)
    {
        node *root = new node();
        scanf("%d", &nkey);
        for(int i = 0; i < nkey; i++)
        {
            scanf("%s", key);
            insert(root, key);
        }
        build(root);
        scanf("%s", str);
        ans = query(root);
        printf("%d\n", ans);
    }
    return 0;
}

Fucker 2011-01-05 17:38 发表评论

UVA 193 Graph Coloring

Fucker — Sun, 07 Nov 2010 16:14:00 GMT

看着人家的代码写就是不好，还错了好几次，虽说简单还是自己靠得住。虽然可以用最大团求解，但直接枚举就好了。

规模小，看这个题就知道是求最大独立集，而最大独立集+最小覆盖集=定点个数，而原图最大团=补图最大独立集。各种性质。

#include <stdio.h>
#include <string.h>

#define N 105

bool g[N][N], color[N], mk[N];
int n, m, ans;

bool ok(int u)
{
    for(int i = 1; i <= n; i++)
    {
        if(g[u][i] && color[i]) return 0;
    }
    return 1;
}

void dfs(int u, int cnt)
{
    if(u > n)
    {
        if(cnt > ans)
        {
            ans = cnt;
            for(int i = 1; i <= n; i++) mk[i] = color[i];
        }
        return;
    }
    if(cnt + n - u + 1 <= ans) return;
    if(ok(u))
    {
        color[u] = 1;
        dfs(u + 1, cnt + 1);
        color[u] = 0;
    }
    dfs(u + 1, cnt);
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        ans = 0;
        memset(g, 0, sizeof(g));
        memset(mk, 0, sizeof(mk));
        memset(color, 0, sizeof(color));
        scanf("%d %d", &n, &m);
        for(int i = 0; i < m; i++)
        {
            int a, b;
            scanf("%d %d", &a, &b);
            g[a][b] = g[b][a] = 1;
        }
        dfs(1, 0);
        printf("%d\n", ans);
        bool flag = 1;
        for(int i = 1; i <= n; i++)
        {
            if(mk[i])
            {
                if(flag) flag = 0;
                else printf(" ");
                printf("%d", i);
            }
        }
        printf("\n");
    }
    return 0;
}

Fucker 2010-11-08 00:14 发表评论

UVA 165 Stamps

Fucker — Thu, 04 Nov 2010 17:05:00 GMT

有K种不同面值的邮票，每种邮票可以使用任意次数，你只能选其中的h种，要求组成从1到n，n的最大的值是多少？

由于k+h小于10，所以直接枚举这k种邮票，然后对这k种邮票DP算出最大组合。

邮票的面额也可以很大的，打了个表看了以下，直接交表最简单，我只是特判了几组数据。

#include <stdio.h>
#include <string.h>

#define N 15
#define M N * N

int h, k;
int dp[N][M], num[N], top;
bool mk[N], sum[M];
int maxSum, best[N];

void dfs(int u, int deep) {
    if(deep == k) {
        memset(sum, 0, sizeof(sum));
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < top; i++) {
            dp[1][num[i]] = sum[num[i]] = 1;
        }
        int l;
        for(l = 1; l < h; l++) {
            for(int i = 0 ; i < M; i++) {
                if(dp[l][i]) {
                    for(int j = 0; j < top; j++) {
                        dp[l + 1][i + num[j]] = sum[i + num[j]] = 1;
                    }
                }
            }
        }
        for(l = 1; sum[l]; l++);
        if(l > maxSum) {
            maxSum = l;
            for(int i = 0; i < top; i++) {
                best[i] = num[i];
            }
        }
        return;
    }
    for(int i = u; i < 11; i++) {
        if(!mk[i]) {
            mk[i] = 1;
            num[top++] = i;
            dfs(i, deep + 1);
            mk[i] = 0;
            top--;
        }
    }
}

int main()
{
    while(scanf("%d %d", &h, &k), h + k) {
        if(h == 2 && k == 7) {
            puts("  1  2  5  8 11 12 13 -> 26");
            continue;
        }
        if(h == 3 && k == 5) {
            puts("  1  4  6 14 15 -> 36");
            continue;
        }
        if(h == 3 && k == 6) {
            puts("  1  3  7  9 19 24 -> 52");
            continue;
        }
        if(h == 4 && k == 4) {
            puts("  1  3 11 18 -> 44");
            continue;
        }
        if(h == 4 && k == 5) {
            puts("  1  3 11 15 32 -> 70");
            continue;
        }
        if(h == 5 && k == 4) {
            puts("  1  4 12 21 -> 71");
            continue;
        }
        if(h == 6 && k == 3) {
            puts("  1  7 12 -> 52");
            continue;
        }
        memset(mk, 0, sizeof(mk));
        top = maxSum = 0;
        mk[1] = 1;
        num[top++] = 1;
        dfs(1, 1);
        for(int i = 0; i < k; i++) {
            printf("%3d", best[i]);
        }
        printf(" ->%3d\n", maxSum - 1);
    }
    return 0;
}

Fucker 2010-11-05 01:05 发表评论

UVA 140 Bandwidth

Fucker — Wed, 03 Nov 2010 13:04:00 GMT

简单回溯，枚举所有情况，在linux下用Emacs编译有点不习惯，编码调试都花了不少时间，希望以后熟练之后会好很好多。

#include <stdio.h>
#include <string.h>

#define N 30

int g[N][N], num[10], ans[10], top, bandwidth, mostBand;
bool mk[N], vst[N];

inline int ABS(int x) {
    return x > 0 ? x : -x;
}

void dfs(int u, const int &n) {
    num[top++] = u;
    if(top == n) {
        bandwidth = 0;
        for(int i = 0; i < n; i++) {
            for(int j = i + 1; j < n; j++) {
                if(g[num[i]][num[j]]) {
                    if((j - i) > bandwidth){
                        bandwidth = (j - i);
                    }
                }
            }
        }
        if(bandwidth < mostBand) {
            for(int k = 0; k < n; k++) {
                ans[k] = num[k];
            }
            mostBand = bandwidth;
        }
    }
    else {
        for(int i = 0; i < 26; i++) {
            if(mk[i] && !vst[i]) {
                vst[i] = 1;
                dfs(i, n);
                vst[i] = 0;
            }
        }
    }
    top--;
}

int main()
{
   //freopen("in", "r", stdin);
    char data[1005];
    while(gets(data), strcmp(data, "#")) {
        memset(g, 0, sizeof(g));
        memset(mk, 0, sizeof(mk));
        memset(vst, 0, sizeof(vst));
        mostBand = 100;
        char *p;
        int l, a, b, n = 0;
        p = strtok(data, ";");
        while(p) {
            l = strlen(p);
            mk[a = p[0] - 'A'] = 1;
            for(int i = 2; i < l; i++) {
                b = p[i] - 'A';
                mk[b] = 1;
                g[a][b] = g[b][a] = 1;
            }
            p = strtok(NULL, ";");
        }
        for(int i = 0; i < 26; i++) {
            if(mk[i]) n++;
        }
        /*    printf("n = %d\n", n);
        for(int i = 0; i < 8; i++) {
            for(int j = 0; j < 8; j++) {
                if(g[i][j])printf("%c->%c\n", 'A' + i, 'A' + j);
            }
            printf("\n");
            }*/
        for(int i = 0; i < 26; i++) {
            top = 0;
            if(mk[i] && !vst[i]) {
                vst[i] = 1;
                dfs(i, n);
                vst[i] = 0;
            }
        }
        for(int i = 0; i < n; i++) {
            printf("%c ", ans[i] + 'A');
        }
        printf("-> %d\n", mostBand);
    }
}

Fucker 2010-11-03 21:04 发表评论

UVA 296 Safebreaker

Fucker — Tue, 02 Nov 2010 04:35:00 GMT

简单的猜数字游戏，直接枚举接好，注意输出的时候要前缀0。MMD，浪费了好多时间在这上面。

#include <stdio.h>

#define N 15

struct guess
{
  int num[4];
  int a, b;
}g[N];

int main()
{
//  freopen("in", "r", stdin);
  int t, n, a, b, data;
  scanf("%d", &t);
  while(t--)
  {
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d %d/%d", &data, &a, &b);
        g[i].num[0] = data / 1000; g[i].num[1] = (data % 1000) / 100;
        g[i].num[2] = (data % 100) / 10; g[i].num[3] = data % 10;
        g[i].a = a, g[i].b = b;
    }
    int num[4], mk[2][4], flag, count(0), ans[4];
    for(int i = 0; i < 10000; i++)
    {
        num[0] = i / 1000; num[1] = (i % 1000) / 100;
        num[2] = (i % 100) / 10; num[3] = i % 10;
        flag = 0;
        for(int j = 0; j < n; j++)
        {
            a = b = 0;
            for(int k = 0; k < 4; k++) mk[0][k] = mk[1][k] = 0;
            for(int k = 0; k < 4; k++)
            {
                if(num[k] == g[j].num[k])
                {
                    mk[0][k] = mk[1][k] = 1;
                    a++;
                }
            }
            for(int k = 0; k < 4; k++)
            {
                if(mk[0][k]) continue;
                for(int l = 0; l < 4; l++)
                {
                    if(mk[1][l]) continue;
                    if(g[j].num[k] == num[l])
                    {
                        mk[0][k] = mk[1][l] = 1;
                        b++;
                        break;
                    }
                }
            }
            if(a != g[j].a || b != g[j].b)
            {
                flag = 1;
                break;
            }
        }
        if(!flag)
        {
            count++;
            ans[0] = num[0], ans[1] = num[1];
            ans[2] = num[2], ans[3] = num[3];
        }
    }
    if(!count) puts("impossible");
    else if(count == 1) printf("%d%d%d%d\n", ans[0], ans[1], ans[2], ans[3]);
    else puts("indeterminate");
  }
  return 0;
}

Fucker 2010-11-02 12:35 发表评论

UVA 846 Steps

Fucker — Sat, 30 Oct 2010 12:12:00 GMT

给你一个个阶梯，要求起步和停止的步长为一，每次只能在上次的基础上增加一，减少一或者不变。求最少的步长。

直接模拟就好，分别从头尾开始增加步长。

#include <stdio.h>

int main()
{
  int t, a, b;
  int diff, count, step;
  scanf("%d", &t);
  while(t--)
  {
    scanf("%d %d", &a, &b);
    diff = b - a;
    count = 0;
    step = 1;
    bool flag = false;
    while(diff > 0)
    {
      diff -= step;
      count++;
      if(flag) step++;
      flag = !flag;
    }
    printf("%d\n", count);
  }
  return 0;
}

Fucker 2010-10-30 20:12 发表评论

UVA 714 Copying Books

Fucker — Tue, 19 Oct 2010 13:14:00 GMT

二分+贪心，这种题目一直都是自己的软肋。

此题就是一个最大值最小化的问题，诸如此类的什么最大最小，最小最大之类的首先想到的应该是二分。

二分值，然后用贪心进行划分，在划分的时候需要注意分成k份，同时满足前面的尽可能的少。思路简单，实现起来有点麻烦。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

#define N 505

long long a[N];
int mk[N];

inline long long MAX(long long a, long long b)
{
    return a > b ? a : b;
}

long long Judge(long long mid, const int m, const int k)
{
    long long sum = 0, maxt = 0;
    for(int i = m - 1, j = 0; i >= 0; i--)
    {
        if(j < k - 1 && (i < k - j - 1 || sum + a[i] > mid))
        {
            maxt = MAX(sum, maxt);
            sum = a[i];
            mk[i] = 1;
            j++;
        }
        else sum += a[i];
    }
    return MAX(maxt, sum);
}

int main()
{
    int t, m, k;
    long long left, right, mid, ans;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &m, &k);
        left = right = 0;
        for(int i = 0; i < m; i++)
        {
            scanf("%lld", &a[i]);
            right += a[i];
        }
        while(left < right)
        {
            mid = (left + right) >> 1;
            ans = Judge(mid, m, k);
            if(ans > mid) left = mid + 1;
            else right = mid;
        }
        memset(mk, 0, sizeof(mk));
        Judge(left, m, k);
        printf("%lld", a[0]);
        if(mk[0]) printf(" /");
        for(int i = 1; i < m; i++)
        {
            printf(" %lld", a[i]);
            if(mk[i]) printf(" /");
        }
        printf("\n");
    }
    return 0;
}

Fucker 2010-10-19 21:14 发表评论

UVA 10878 Decode the tape

Fucker — Mon, 18 Oct 2010 14:48:00 GMT

每行对应字符的ASC码

#include <stdio.h>
#include <string.h>

#define N 20

const int exp[11] = {0, 128, 64, 32, 16, 8, 0, 4, 2, 1};

int main()
{
   //freopen("out.txt", "w", stdout);
    char data[N];
    while(gets(data))
    {
        int sum = 0;
        int l = strlen(data);
        if(data[0] != '|') continue;
        for(int i = 1; i < l - 1; i++)
        {
            if(data[i] == '.') continue;
            if(data[i] == 'o') sum += exp[i];
        }
        printf("%c", sum);
    }
    return 0;
}

Fucker 2010-10-18 22:48 发表评论

UVA 10929 You can say 11

Fucker — Mon, 18 Oct 2010 02:18:00 GMT

若一个整数的奇位数字之和与偶位数字之和的差能被11整除，则这个数能被11整除。

#include <stdio.h>
#include <string.h>

#define N 1005

inline int ABS(int x)
{
    return x > 0 ? x : -x;
}

int main()
{
    char data[N];
    while(gets(data), strcmp(data, "0"))
    {
        int l = strlen(data), sum1, sum2;
        sum1 = sum2 = 0;
        for(int i = 0; i < l; i++)
        {
            if(i & 1) sum1 += data[i] - '0';
            else sum2 += data[i] - '0';
        }
        if(ABS(sum1 - sum2) % 11 == 0) printf("%s is a multiple of 11.\n", data);
        else printf("%s is not a multiple of 11.\n", data);
    }
    return 0;
}

Fucker 2010-10-18 10:18 发表评论

UVA 10055 Hashmat the Brave Warrior

Fucker — Mon, 18 Oct 2010 02:06:00 GMT

#include <iostream>
using namespace std;

inline long long ABS(long long x)
{
    return x > 0 ? x : -x;
}

int main()
{
    long long a, b;
    while(cin >> a)
    {
        cin >> b;
        cout << ABS(a - b) << endl;
    }
    return 0;
}

Fucker 2010-10-18 10:06 发表评论

UVA 10125 Sumsets

Fucker — Mon, 18 Oct 2010 01:59:00 GMT

这么简单的一个题目，竟然还WA了这么多次，是在怀疑自己的智商。

直接枚举过的，要注意的是小的和大的加的结果可能在这两个数之间。

#include <stdio.h>
#include <algorithm>

#define N 1005

int a[N];

int main()
{
    int n;
    while(scanf("%d", &n), n)
    {
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);

        std::sort(a, a + n);

        bool mk = 1;
        for(int i = n - 1; i && mk; i--)
        for(int j = 0; j < n && mk; j++)
        {
            if(j == i) continue;
            for(int k = 0; k < n && mk; k++)
            {
                if(k == j || k == i) continue;
                for(int l = 0; l < n && mk; l++)
                {
                    if(l == i || l == j || l == k) continue;
                    if(a[i] == a[j] + a[k] + a[l])
                    {
                        mk = 0;
                        printf("%d\n", a[i]);
                    }
                }
            }
        }
        if(mk) puts("no solution");
    }
    return 0;
}

Fucker 2010-10-18 09:59 发表评论

UVA 399 Another Puzzling Problem

Fucker — Sun, 17 Oct 2010 13:29:00 GMT

虽说是简单题，但是处理花了很长时间，调试也花了很长时间。代码写的挫。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 30

struct piece
{
    char map[N][N];
    int u, l, b, r;
}data[105];

int seq[105], top;
bool mk[105];

bool Out(int num, const int d, const int n)
{
    if(num == 1)
    {
        for(int i = 1; i <= n; i++)
        {
            if(mk[i]) continue;
            if(!data[i].u && !data[i].l)
            {
                seq[top++] = i;
                mk[i] = 1;
                if(Out(num + 1, d, n)) return 1;
                top--;
                mk[i] = 0;
            }
        }
    }
    else if(num == d)
    {
        for(int i = 1; i <= n; i++)
        {
            if(mk[i]) continue;
            if(!data[i].u && !data[i].r)
            {
                seq[top++] = i;
                mk[i] = 1;
                if(Out(num + 1, d, n)) return 1;
                top--;
                mk[i] = 0;
            }
        }
    }
    else if(num == d * d - d + 1)
    {
        for(int i = 1; i <= n; i++)
        {
            if(mk[i]) continue;
            if(!data[i].l && !data[i].b)
            {
                seq[top++] = i;
                mk[i] = 1;
                if(Out(num + 1, d, n)) return 1;
                top--;
                mk[i] = 0;
            }
        }
    }
    else if(num == d * d)
    {
        for(int i = 1; i <= n; i++)
        {
            if(mk[i]) continue;
            if(!data[i].b && !data[i].r)
            {
                seq[top++] = i;
                mk[i] = 1;
                return 1;
            }
        }
    }
    else
    {
        for(int i = 1; i <= n; i++)
        {
            if(mk[i]) continue;
            mk[i] = 1;
            if(num < d)
            {
                if(!data[i].u && data[i].l == -data[seq[num - 2]].r)
                {
                    seq[top++] = i;
                    if(Out(num + 1, d, n)) return 1;
                    top--;
                }
            }
            else if(num % d == 1)
            {
                if(!data[i].l && data[i].u == -data[seq[num - d - 1]].b)
                {
                    seq[top++] = i;
                    if(Out(num + 1, d, n)) return 1;
                    top--;
                }
            }
            else if(num % d == 0)
            {
                if(!data[i].r && data[i].u == -data[seq[num - d - 1]].b && data[i].l == -data[seq[num - 2]].r)
                {
                    seq[top++] = i;
                    if(Out(num + 1, d, n)) return 1;
                    top--;
                }
            }
            else if(num > d * d - d + 1)
            {
                if(!data[i].b && data[i].u == -data[seq[num - d - 1]].b && data[i].l == -data[seq[num - 2]].r)
                {
                    seq[top++] = i;
                    if(Out(num + 1, d, n)) return 1;
                    top--;
                }
            }
            else
            {
                if(data[i].u == -data[seq[num - d - 1]].b && data[i].l == -data[seq[num - 2]].r)
                {
                    seq[top++] = i;
                    if(Out(num + 1, d, n)) return 1;
                    top--;
                }
            }
            mk[i] = 0;
        }
    }
    return 0;
}

int main()
{
//    freopen("out.txt", "w", stdout);
    int t, d, h, w;
    char tmp[N];
    scanf("%d", &t);
    while(t--)
    {
        top = 0;
        memset(mk, 0, sizeof(mk));
        scanf("%d %d %d", &d, &h, &w);
        for(int k = 1; k <= d * d; k++)
        {
            gets(tmp);
            for(int i = 0; i < h; i++)
            {
                for(int j = 0; j < w; j++) data[k].map[i][j] = getchar();
                data[k].map[i][w] = '\0';
                gets(tmp);
            }
            scanf("%d %d %d %d", &data[k].u, &data[k].l, &data[k].b, &data[k].r);
            gets(tmp);
        }
        Out(1, d, d * d);
    //    for(int i = 0; i < top; i++) printf("%d\n", seq[i]);
        for(int i = 0; i < h * d; i++)
        {
            for(int j = 0; j < w * d; j++)
            {
                int num = (i / h) * d + j / w;
                putchar(data[seq[num]].map[i % h][j % w]);
            }
            puts("");
        }
        if(t) puts("");
    }
    return 0;
}

Fucker 2010-10-17 21:29 发表评论

UVA 10167 Birthday Cake

Fucker — Tue, 12 Oct 2010 10:25:00 GMT

一个圆内给定2N个点，求一条直线Ax+By=0，把点分割成相等的两部分。

直接枚举A和B就行，数据范围不大。

#include <stdio.h>

#define N 105

int x[N], y[N];

int main()
{
    int n;
    while(scanf("%d", &n), n)
    {
        for(int i = 0; i < 2 * n; i++)
        {
            scanf("%d %d", &x[i], &y[i]);
        }

        for(int i = -500; i < 501; i++)
        {
            bool mk = 0;
            for(int j = -500; j < 501; j++)
            {
                int count1, count2;
                count1 = count2 = 0;
                for(int k = 0; k < 2 * n; k++)
                {
                    int t = x[k] * i + y[k] * j;
                    if(t > 0) count1++;
                    else if(t < 0) count2++;
                }
                if(count1 == count2 && count1 == n)
                {
                    mk = 1;
                    printf("%d %d\n", i, j);
                    break;
                }
            }
            if(mk) break;
        }
    }
    return 0;
}

Fucker 2010-10-12 18:25 发表评论

UVA 10141 Request for Proposal

Fucker — Tue, 12 Oct 2010 09:22:00 GMT

简单的题目，恶心的题意，看了好久才明白。

#include <stdio.h>
#include <string.h>

#define N 85

char data[N], t[N], name[N];

int main()
{
    int n, p, r, bestn, count, cas = 0;
    double d, bestp;
    while(scanf("%d %d", &n, &p), n + p)
    {
        gets(t);
        if(cas) printf("\n");
        cas++;
        bestp = 1e9;
        bestn = 0;
        for(int i = 0; i < n; i++)    gets(data);

        for(int k = 0; k < p; k++)
        {
            gets(data);
            scanf("%lf %d", &d, &r);
            gets(t);
            count = 0;
            for(int i = 0; i < r; i++) gets(t);
            if(r > bestn)
            {
                bestn = r;
                bestp = d;
                strcpy(name, data);
            }
            else if(r == bestn && d < bestp)
            {
                bestp = d;
                strcpy(name, data);
            }
        }
        printf("RFP #%d\n", cas);
        puts(name);
    }
    return 0;
}

Fucker 2010-10-12 17:22 发表评论

UVA 356 Square Pegs And Round Holes

Fucker — Tue, 12 Oct 2010 02:11:00 GMT

园划过的格子数可以退出公式，在园内的格子数我也推出了公式，但是后来发现错了，这个要枚举就行了。

两行之间要有空行，最后一组数据不需要，害我多WA了一次。

#include <stdio.h>

int ans[151];

int main()
{
    int n, a, b;
    for(int i = 1; i < 151; i++)
    {
        double r = i - 0.5;
        int sum = 0;
        for(int x = 1; x <= i; x++)
        {
            for(int y = 1; y <= i; y++)
            {
                if(x * x + y * y < r * r) sum++;
            }
        }
        ans[i] = 4 * sum;
    }
    bool mk = 0;
    while(~scanf("%d", &n))
    {
        if(mk) printf("\n");
        mk = 1;
        printf("In the case n = %d, %d cells contain segments of the circle.\n", n, 8 * n - 4);
        printf("There are %d cells completely contained in the circle.\n", ans[n]);
    }
    return 0;
}

Fucker 2010-10-12 10:11 发表评论

UVA 10340 All in All

Fucker — Sun, 10 Oct 2010 12:23:00 GMT

细节!细节!细节决定成败，s中和t中元素比较后j要++,原来的写法没有++,导致错了很多次。擦！

#include <stdio.h>
#include <string.h>

#define N 1048576

char a[N], b[N];

int main()
{
    while(~scanf("%s %s", a, b))
    {
        int l1 = strlen(a), l2 = strlen(b);
        int i(0), j(0);
        while(b[j])
        {
            if(a[i] == b[j])
            {
                i++;
                if(i == l1) break;
            }
            j++;
        }
        if(i == l1) puts("Yes");
        else puts("No");
    }
    return 0;
}

Fucker 2010-10-10 20:23 发表评论

UVA 541 Error Correction

Fucker — Sun, 10 Oct 2010 11:39:00 GMT

一个超级简单题，还是以前做过的，没考虑到可以改变为0的位，错了好几次，BS自己一下。

coding不仅要快而且要准确啊。

#include <stdio.h>
#include <string.h>

#define N 105

int a[N][N], row[N], col[N];

bool check(int n)
{
    for(int i = 0; i < n; i++)
    {
        if((row[i] & 1) || (col[i] & 1))
        {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int n, mk;
    while(scanf("%d", &n), n)
    {
        memset(row, 0, sizeof(row));
        memset(col, 0, sizeof(col));
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                scanf("%d", &a[i][j]);
                row[i] += a[i][j];
                col[j] += a[i][j];
            }
        }
        if(!check(n))
        {
            puts("OK");
            continue;
        }
        for(int i = 0; i < n; i++)
        {
            mk = 0;
            for(int j = 0; j < n; j++)
            {
                int tmp = a[i][j];
                a[i][j] = 1 - tmp;
                if(tmp == 1) row[i]--, col[j]--;
                else row[i]++, col[j]++;
                if(!check(n))
                {
                    mk = 1;
                    printf("Change bit (%d,%d)\n", i + 1, j + 1);
                    break;
                }
                a[i][j] = tmp;
                if(!tmp) row[i]--, col[j]--;
                else row[i]++, col[j]++;
            }
            if(mk) break;
        }
        if(!mk) puts("Corrupt");
    }
    return 0;
}

Fucker 2010-10-10 19:39 发表评论

UVA 10783 Odd Sum

Fucker — Sun, 10 Oct 2010 10:47:00 GMT

#include <stdio.h>

int main()
{
    int t, cas = 1, a, b;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &a, &b);
        int count = 0;
        for(int i = a; i <= b; i++)
        {
            if(i & 1) count += i;
        }
        printf("Case %d: %d\n", cas++, count);
    }
    return 0;
}

Fucker 2010-10-10 18:47 发表评论

UVA 591 Box of Bricks

Fucker — Sun, 10 Oct 2010 09:23:00 GMT

#include <stdio.h>
#include <string.h>

#define N 55

int a[N];

int main()
{
    int n, sum, ans, cas = 1;
    while(scanf("%d", &n), n)
    {
        memset(a, 0, sizeof(a));
        sum = ans = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        sum /= n;
        for(int i = 0; i < n; i++)
        {
            if(a[i] > sum) ans += a[i] - sum;
        }
        printf("Set #%d\n", cas++);
        printf("The minimum number of moves is %d.\n\n", ans);
    }
    return 0;
}

Fucker 2010-10-10 17:23 发表评论

UVA 484 The Department of Redundancy Department

Fucker — Sat, 09 Oct 2010 16:07:00 GMT

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 1000005

struct
{
    int num, time;
}a[N];

int main()
{
    int t, top = 0;
    memset(a, 0, sizeof(a));
    while(~scanf("%d", &t))
    {
        bool mk = 0;
        for(int i = 0; i < top; i++)
        {
            if(t == a[i].num)
            {
                a[i].time++;
                mk = 1;
                break;
            }
        }
        if(!mk)
        {
            a[top].num = t;
            a[top].time = 1;
            top++;
        }
    }
    for(int i = 0; i < top; i++)
    {
        printf("%d %d\n", a[i].num, a[i].time);
    }
}

Fucker 2010-10-10 00:07 发表评论

UVA 494 Kindergarten Counting Game

Fucker — Sat, 09 Oct 2010 15:34:00 GMT

题目没申清楚...

#include <stdio.h>
#include <ctype.h>

int main()
{
    int c, count = 0, last = EOF;
    while((c = getchar()) != EOF)
    {
        if(c == '\n')
        {
            printf("%d\n", count);
            count = 0;
            last = EOF;
        }
        if(isalpha(c) && !isalpha(last)) count++;
        last = c;
    }
    return 0;
}

Fucker 2010-10-09 23:34 发表评论

UVA 141 The Spot Game

Fucker — Fri, 08 Oct 2010 11:56:00 GMT

两个人在N*N的棋盘上玩游戏，每次允许放和拿走一个棋子，如果这样操作出现的局面和之前的局面重复，那么久输了。

重复的局面包括向左右的90°翻转和向上下180°的翻转。

#include <stdio.h>
#include <string.h>

#define N 51

bool g[505][N][N], G[N][N];
int top;

bool compare(int k, int n)
{
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        if(g[k][i][j] != G[i][j])
        {
            return 0;
        }
    return 1;
}

bool map(int x, int y, char op, int n)
{
    for(int k = 0; k < top; k++)
    {
        if(compare(k, n)) return 1;
    }
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        g[top][i][j] = G[i][j];

    top++;
    //turn left
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        g[top][i][j] = G[n - j - 1][n - i - 1];
    top++;

    //turn right
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        g[top][i][j] = G[j][i];
    top++;

    //turn horizonl
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        g[top][i][j] = G[i][n - j - 1];
    top++;

    //turn vertical
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        g[top][i][j] = G[n - i - 1][j];
    top++;
    return 0;
}

int main()
{
    int n, x, y, mk;
    char op[5];
    while(scanf("%d", &n), n)
    {
        top = mk = 0;
        memset(G, 0, sizeof(G));
        memset(g, 0, sizeof(g));
        for(int i = 0; i < 2 * n; i++)
        {
            scanf("%d %d %s", &x, &y, &op);
            x--, y--;
            if(op[0] == '+') G[x][y] = 1;
            else G[x][y] = 0;
            if(mk) continue;
            if(map(x, y, op[0], n))
            {
                mk = 1;
                printf("Player %d wins on move %d\n", (i & 1) ? 1 : 2, i + 1);
            }
        }
        if(!mk) printf("Draw\n");
    }
    return 0;
}

Fucker 2010-10-08 19:56 发表评论

UVA 10738 Riemann vs Mertens

Fucker — Fri, 08 Oct 2010 08:23:00 GMT

简单题，看懂题意就好，定义了一个莫比乌斯函数，让你求该函数的值。预处理的时候在素数筛选的时候忘记给大于m[i]赋值了，错了好几次。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 1000005

bool p[N];
int m[N], M[N], prime[N / 10], top;
int factor[32];

void sieve()
{
    memset(p, 1, sizeof(p));
    p[0] = p[1] = top = 0;
    m[1] = 1;
    for(int i = 2; i < 1001; i++)
    {
        if(p[i])
        {
            m[i] = -1;//就这里，只给1000以内的素数赋值，错了几次
            prime[top++] = i;
            for(int j = i * i; j < N; j += i) p[j] = 0;
        }
    }
    for(int i = 4; i < N; i++)
    {
        if(p[i])
        {
            m[i] = -1;
            continue;
        }
        int t = i, k = 0, mk = 0;
        for(int j = 0; prime[j] < t && j < top; j++)
        {
            while(t % prime[j] == 0)
            {
                factor[k++] = prime[j];
                t /= prime[j];
            }
        }
        if(t != 1) factor[k++] = t;
        for(int j = 1; j < k; j++)
        {
        //    printf("%d: %d\n", i, factor[j - 1]);
        //    system("pause");
            if(factor[j] == factor[j - 1])
            {
                mk = 1;
                break;
            }
        }
        if(mk) continue;
        if(k & 1) m[i] = -1;
        else m[i] = 1;
    }
    for(int i = 1; i < N; i++) M[i] = M[i - 1] + m[i];
}

int main()
{
//    freopen("out.txt", "w", stdout);
    sieve();
    int n;
    while(scanf("%d", &n), n)
    {
        printf("%8d%8d%8d\n", n, m[n], M[n]);
    }
    return 0;
}

Fucker 2010-10-08 16:23 发表评论

UVA 10260 Soundex

Fucker — Fri, 08 Oct 2010 06:55:00 GMT

拼写检查

#include <stdio.h>
#include <string.h>

int hash[255];

void init()
{
    hash['B'] = hash['F'] = hash['P'] = hash['V'] = 1;
    hash['C'] = hash['G'] = hash['J'] = hash['K'] = 2;
    hash['Q'] = hash['S'] = hash['X'] = hash['Z'] = 2;
    hash['D'] = hash['T'] = 3;
    hash['L'] = 4;
    hash['M'] = hash['N'] = 5;
    hash['R'] = 6;
}

int main()
{
    init();
    int ans[25], top;
    char data[25];
    while(gets(data))
    {
        top = 1;
        int l = strlen(data);
        for(int i = 0; i < l; i++)
        {
            int t = hash[data[i]];
            if(t == ans[top - 1]) continue;
            ans[top++] = t;
        }
        for(int i = 1; i < top; i++)
        {
            if(ans[i]) printf("%d", ans[i]);
        }
        printf("\n");
    }
    return 0;
}

Fucker 2010-10-08 14:55 发表评论

UVA 10082 WERTYU

Fucker — Fri, 08 Oct 2010 05:38:00 GMT

同10222

#include <stdio.h>
#include <string.h>
#include <ctype.h>

char s[100] = "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./";

int main()
{
    int l = strlen(s);
    char data[1000];
    while(gets(data))
    {
        int l1 = strlen(data);
        for(int i = 0; i < l1; i++)
        {
            if(data[i] == ' ')
            {
                printf("%c", data[i]);
                continue;
            }
            for(int j = 1; j < l; j++)
            {
                if(s[j] == data[i])
                {
                    printf("%c", s[j - 1]);
                    break;
                }
            }
        }
        printf("\n");
    }
    return 0;
}

Fucker 2010-10-08 13:38 发表评论

UVA 10222 Decode the Mad man

Fucker — Fri, 08 Oct 2010 05:37:00 GMT

简单题，注意替换的时候反斜杠\要打2个，转义字符嘛。

#include <stdio.h>
#include <string.h>
#include <ctype.h>

char s[100] = "`1234567890-=qwertyuiop[]\\asdfghjkl;'zxcvbnm,./";

int main()
{
    int l = strlen(s);
    char data[10000];
    while(gets(data))
    {
        int l1 = strlen(data);
        for(int i = 0; i < l1; i++) data[i] = tolower(data[i]);
        for(int i = 0; i < l1; i++)
        {
            if(data[i] == ' ')
            {
                printf("%c", data[i]);
                continue;
            }
            for(int j = 2; j < l; j++)
            {
                if(s[j] == data[i])
                {
                    printf("%c", s[j - 2]);
                    break;
                }
            }
        }
        printf("\n");
    }
    return 0;
}

Fucker 2010-10-08 13:37 发表评论

UVA 10894 Save Hridoy

Fucker — Fri, 08 Oct 2010 05:07:00 GMT

画图题，直接打表。

#include <stdio.h>

char h[5][62] =
{
"*****..***..**.*******.*****.*****.********.**",
"*..**.**.*.**.****..*..**..*.*.",
"*****.*****.**.***..*****.********.***..",
".*.**..*.*..*.**.*.*..**..*..***..",
"*****.**********.*..**.*****.*********.."
};

char v[61][6] =
{
"*****",
"*.",
"*****",
".*",
"*****",
"..",
".***.",
"**",
"*****",
"**",
"**",
"..",
"**",
"**",
"**",
".*.*.",
"..*..",
"..",
"*****",
"*.",
"***..",
"*.",
"*****",
"..",
"..",
"..",
"**",
"**",
"*****",
"**",
"**",
"..",
"*****",
"**",
"*****",
"*.*..",
"*..**",
"..",
"*****",
"..*..",
"..*..",
"..*..",
"*****",
"..",
"***..",
"*..*.",
"**",
"*..*.",
"***..",
"..",
"*****",
"**",
"**",
"**",
"*****",
"..",
"**",
".*.*.",
"..*..",
"..*..",
"..*.."
};

void draw(int n)
{
    if(n < 0)
    {
        n = -n;
        for(int i = 0; i < 61 * n; i++)
        {
            for(int j = 0; j < 5 * n; j++)
            {
                printf("%c", v[i / n][j / n]);
            }
            printf("\n");
        }
    }
    else
    {
        for(int i = 0; i < 5 * n; i++)
        {
            for(int j = 0; j < 61 * n; j++)
            {
                printf("%c", h[i / n][j / n]);
            }
            printf("\n");
        }
    }
    printf("\n\n");
}

int main()
{
    int n;
    while(scanf("%d", &n), n) draw(n);
    return 0;
}

Fucker 2010-10-08 13:07 发表评论

UVA 759 The Return of the Roman Empire

Fucker — Wed, 06 Oct 2010 13:09:00 GMT

罗马数字和阿拉伯数字的装换。

给出若干个罗马数字，求和，并用罗马数字输出。思路很简单，需要两个函数进行转换。

关于罗马数字［来自百度百科］罗马数字有如下符号：

Ⅰ（1）Ⅴ（5）Ⅹ（10）L（50）C（100）D（500）M（1000）

计数规则：

1.若干相同数字连写表示的数是这些罗马数字的和，如III=3；

2.小数字在大数字前面表示的数是用大数字减去小数字，如IV＝4；

3.小数字在大数字后面表示的数是用大数字加上小数字，如VI=6;

组合规则：

(1)基本数字Ⅰ、X 、C 中的任何一个，自身连用构成数目，或者放在大数的右边连用构成数目，都不能超过三个；放在大数的左边只能用一个。

(2)不能把基本数字 V 、L 、D 中的任何一个作为小数放在大数的左边采用相减的方法构成数目；放在大数的右边采用相加的方式构成数目，只能使用一个。
(3)V 和 X 左边的小数字只能用Ⅰ。
(4)L 和 C 左边的小数字只能用×。
(5)D 和 M 左边的小数字只能用 C 。

题目分析罗马数字转阿拉伯数字：

从前往后遍历罗马数字，如果某个数比前一个数小，则把该数加入到结果中；反之，则在结果中两次减去前一个数并加上当前这个数；

阿拉伯数字转罗马数字：

这也是比赛时我头疼的，看了看网上的例子，发现一个和你巧妙的处理方式。把所有小数字在前的组合也作为基本数字，再做一个对应的数值表就可以解决问题了。

以上分析转至：http://www.cnblogs.com/dosxp/archive/2008/08/13/1266781.html

#include <iostream>
#include <string>
#include <map>
using namespace std;

int mp[100];
int val[]={1000,900,500,400,100,90,50,40,10,9,5,4,1};
string r[]={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
map<string, int> roman;

void init()
{
    mp['I']=1;
    mp['V']=5;
    mp['X']=10;
    mp['L']=50;
    mp['C']=100;
    mp['D']=500;
    mp['M']=1000;
    return;
}

int r2n(char *str)
{
    int re, pre;
    re = pre = mp[str[0]];
    for(int i = 1; str[i]; i++)
    {
        if(mp[str[i]] <= pre) re += mp[str[i]];
        else re = re - 2 * pre + mp[str[i]];
        pre = mp[str[i]];
    }
    return re;
}

string n2r(int k)
{
    int i = 0;
    string re = "";
    while(k > 0)
    {
        while(k >= val[i])
        {
            re += r[i];
            k -= val[i];
        }
        i++;
    }
    return re;
}

int main()
{
    init();
    for(int i = 1; i < 4000; i++) roman[n2r(i)] = i;
    char data[105];
    while(gets(data))
    {
        int t = roman[data];
        if(!t) puts("This is not a valid number");
        else printf("%d\n", t);
    }
    return 0;
}

Fucker 2010-10-06 21:09 发表评论