C++博客-purplest-随笔分类-模拟

UVa 297 Quadtrees

purplest — Mon, 19 Mar 2012 09:26:00 GMT

哎，太水了，递归建树写了好久的说，太弱了。。。

就是建个四叉树，然后统计占了多少像素，2Y，第一次犯2数组开小了送了个RE。。。

貌似还可以直接建成满四叉树，不过算晕了，没推出子节点咋算的。。真的太弱了= =

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

struct Node
{
    int flag, child[4];
};

int ncase, sz, res;
char str1[2000], str2[2000];
Node treea[4500], treeb[4500];

void build(int left, int right, int fa, char *str, Node *tree)
{
    for ( int i = left, k = 0 ; i < right && k < 4 ; i++, k++ )
    {
        int t = sz++;
        if ( str[i] == 'p' )
        {
            int j, times=4;
            for ( j = i+1 ; j < right && times ; j++, times-- )
                if ( str[j] == 'p' )
                    times+=4;
            tree[t].flag = 1;
            tree[fa].child[k]=t;
            build(i+1, j, t, str, tree);
            i = j-1;
        }
        else
        {
            tree[t].flag = str[i] == 'e' ? 2 :3;
            tree[fa].child[k]=t;
        }
    }
}

void init(char *str, Node *tree)
{
    int len = strlen(str);
    sz = 1;
    build(0, len, 1, str, tree);
}

void dfs(Node *tree, int root, int deep)
{
    if (tree[root].flag==2)
        return;
    if (tree[root].flag==3)
    {
        res+=1<<deep;
        return;
    }
    if (tree[root].flag==1)
        for ( int i = 0 ; i < 4 ; i++ )
            dfs(tree, tree[root].child[i], deep-2);
}

void get(int root1, int root2, int deep)
{
    if ( !root1 && !root2 )
        return;
    if (treea[root1].flag==3 || treeb[root2].flag==3)
        res+=1<<deep;
    if (treea[root1].flag==2||treeb[root2].flag==2)
        treea[root1].flag==2 ? dfs(treea, root1, deep) : dfs(treeb, root2, deep);
    if (treeb[root2].flag==1&&treea[root1].flag==1)
        for ( int i = 0 ; i < 4 ; i++ )
            get(treea[root1].child[i], treeb[root2].child[i], deep-2);
}

int main()
{
    while ( EOF != scanf("%d", &ncase) )
    {
        memset(treea, 0, sizeof(treea));
        memset(treeb, 0, sizeof(treeb));
        scanf("%s %s", str1, str2);
        init(str1, treea);
        init(str2, treeb);
        res = 0;
        get(1,1,10);
        printf("There are %d black pixels.\n", res);
    }
    return 0;
}

purplest 2012-03-19 17:26 发表评论

UVa 548

purplest — Fri, 16 Mar 2012 09:30:00 GMT

题意：给中序，后序，然后还原二叉树，再输出从root到leaf节点中节点和最小的那个leaf

很无语的是用index这个变量，居然被判了个CE。。。

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

struct Node
{
    int l, r, v;
};

char ch;
int inorder[10100], postorder[10100];
Node node[10100];
int sz, minn, indexn;

void recover(int inl, int inr, int postl, int postr, int &fa)
{
    if ( inr<inl || postr<postl )
        return;
    int p = ++sz;
    fa = sz;
    node[p].v=postorder[postr];
    int k;
    for ( k = inl ; k <= inr && inorder[k]!=postorder[postr] ; k++);
    int len = k-inl;
    recover(inl, k-1, postl, postl+len-1, node[p].l);
    recover(k+1, inr, postl+len, postr-1, node[p].r);
}

void dfs(int p, int num)
{
    if (!node[p].l&&!node[p].r)
    {
        if ( node[p].v+num < minn )
        {
            indexn = node[p].v;
            minn= node[p].v+num;
        }
        else if ( node[p].v+num == minn )
            if ( indexn > node[p].v)
                indexn = node[p].v;
        return;
    }
    node[p].v+=num;
    if (node[p].l)
        dfs(node[p].l, node[p].v);
    if (node[p].r)
        dfs(node[p].r, node[p].v);
}

int main()
{
    while ( EOF != scanf("%d%c", inorder, &ch) )
    {
        indexn = minn = ~1u>>1;
        memset(node, 0, sizeof(node));
        int top;
        for ( top = 1 ; ch != '\n' ; top++ )
            scanf("%d%c", inorder+top, &ch);
        for ( int i = 0 ; i < top ; i++ )
            scanf("%d", postorder+i);
        recover(0, top-1, 0, top-1, node[sz=0].r);
        dfs(1, 0);
        printf("%d\n", indexn);
    }
    return 0;
}

purplest 2012-03-16 17:30 发表评论

UVa 112 poj 1145

purplest — Fri, 16 Mar 2012 07:56:00 GMT

这题WA死了，5Y。。。

开始没想到还有负数的可能，送了2个WA，一次是犯2忘关调试了，还有一次是 0 () 这组数据没处理

好久没写模拟了，太没状态了，哎，还好把代码给压缩到了100以内

附上AC代码~

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

struct Node
{
    int l, r, v;
};

Node node[101000];
int num, sum;
char str[2048],stack[101000];
bool flag;

void build(int l, int r, int &fa)
{
    if ( r-l == 1 )
    {
        fa = 0;
        return;
    }
    int t = sum++, j;
    fa = t;
    int len = 0, temp = 0, fh = 1, left = l;
    if ( stack[l+1] == '-' )
    {
        fh=-1;
        left++;
    }
    for ( int i = left+1 ; i < r && ('0' <= stack[i] && stack[i] <= '9' ) ; i++, len++ )
        temp=temp*10+stack[i]-'0';
    node[t].v=temp*fh;
    for ( int i = left+2+len , k = 1 ; i < r && k ; i++ )
    {
        if (stack[i]==')')
            k--;
        else if (stack[i]=='(')
            k++;
        if (!k)
            j = i;
    }
    build(left+len+1, j, node[t].l);
    build(j+1, r-1, node[t].r);
}

void dfs(int p, int n)
{
    node[p].v+=n;
    if (node[p].l)
        dfs(node[p].l, node[p].v);
    if (node[p].r)
        dfs(node[p].r, node[p].v);
    if (!node[p].l&&!node[p].r)
        if (node[p].v==num)
            flag = 1;
}

int main()
{
    while ( EOF != scanf("%d", &num) )
    {
        memset(stack, 0, sizeof(stack));
        memset(node, 0, sizeof(node));
        int top = 0, k = 0;
        for ( int i = 0 ; top == 0 || k ; i = 0)
        {
            gets(str);
            for ( int len = strlen(str) ; i < len ; i++ )
            {
                while ( ' ' == str[i] ) i++;
                if ( '(' == str[i] )
                {
                    k++;
                    stack[top++]=str[i];
                }
                else if( ')' == str[i] )
                {
                    k--;
                    stack[top++]=str[i];
                }
                else if (('0' <= str[i] && str[i] <= '9') || ( '-' == str[i] ))
                    stack[top++]=str[i];
            }
        }
        sum = 1;
        flag = 0;
        build(0, top-1, node[0].r);
        if ( top == 2 )
            puts("no");
        else
        {
            dfs(1, 0);
            flag ? puts("yes"): puts("no");
        }
    }
    return 0;
}

purplest 2012-03-16 15:56 发表评论

poj 1782 Run Length Encoding

purplest — Tue, 23 Aug 2011 16:42:00 GMT

模拟题，trick很多。。。

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

char str[1000];

int main()
{
    memset(str, 0, sizeof(str));
    while ( !feof(stdin) )
    {
        gets(str);
        if (!str[0])
        {
            putchar(10);
            continue;
        }
        int i, j;
        for ( i = 0 ; str[i] ; i++ )
        {
            if (str[i] == str[i+1])
            {
                for ( j = i ; str[i] == str[i+1] && (i-j) < 8 ; i++ );
                printf("%d%c", i-j+1, str[i]);
            }
            else
            {
                printf("1");
                for ( j = i ; str[i] != str[i+1] && str[i] ; i++ )
                {
                    if ('1' == str[i]) printf("1");
                    printf("%c", str[i]);
                }
                printf("1");
                if ( str[i] == str[i+1]) i--;
            }
        }
        putchar(10);
        memset(str, 0, sizeof(str));
    }
    return 0;
}

purplest 2011-08-24 00:42 发表评论

poj 1672 Fully Diversified Sequences of Sets

purplest — Tue, 16 Aug 2011 16:48:00 GMT

要分奇偶数

#include <stdio.h>

int n;

void sing()
{
    int i, j, k;
    for ( i = 1 ; i < n ; i += 2 )
        printf("%d ", i);
    printf("%d\n", n);
    for ( i = 1 ; i < ((n+1)>>1); i++ )
    {
        k = 2;
        while(k--)
        {
            for ( j = 2 * i ; j < n ; j++ )
                printf("%d ", j);
            printf("%d\n", n);
        }
    }
}
void dual()
{
    int i, j, k;
    for ( i = 1 ; i < n ; i++ )
        printf("%d ", i);
    printf("%d\n", n);
    for ( i = 2 ; i < n ; i += 2 )
        printf("%d ", i);
    printf("%d\n", n);
    for ( i = 3 ; i < n ; i += 2 )
    {
        k = 2;
        while ( k-- )
        {
            for ( j = i ; j < n ; j++ )
                printf("%d ", j);
            printf("%d\n", n);
        }
    }
}

int main()
{
    while (scanf("%d", &n), n)
    {
        int total= n*(n+1)/2;
        if ( total & 1 )
            printf("0\n");
        else
        {
            printf("%d\n", n);
            if ( n & 1 )
                sing();
            else
                dual();
        }
        putchar(10);
    }
    return 0;
}

purplest 2011-08-17 00:48 发表评论

poj 2275 Flipping Pancake

purplest — Tue, 16 Aug 2011 16:44:00 GMT

翻饼的问题，纯模拟

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

int data[100], t[100], num[100];
int n;

int cmp(const void *a, const void *b)
{
    return *(int *)a-*(int *)b;
}

void reverse(int m)
{
    int mid=(m+1)>>1, t;
    for ( int i = 0 ; i < mid ; i++ )
    {
        t= data[i];
        data[i]= data[m-i];
        data[m-i]= t;
    }
}

int main()
{
    while ( scanf("%d", &n), n)
    {
        int i, j, p, count= 0;
        for ( i = 0 ; i < n ; i++ )
        {
            scanf("%d", data+i);
            t[i]= data[i];
        }
        qsort(t, n, sizeof(int), cmp);
        for ( p = n-1 ; p >= 0 ; p-- )
        {
            if ( data[p] != t[p] )
            {
                for ( i = 0 ; i < n && t[p] != data[i] ; i++ );
                if ( p > 0 )
                {
                    for ( j = 0 ; j < n && t[p-1] != data[j] ; j++ );
                    if ( j+1 < i )
                    {
                        if ( 0 < j )
                        {
                            reverse(j);
                            num[count++]=j;
                        }
                        reverse(i-1);
                        num[count++]= i-1;
                    }
                }
                if (i)
                {
                    reverse(i);
                    num[count++]= i;
                }
                reverse(p);
                num[count++]= p;
            }
        }
        printf("%d", count);
        for ( i = 0 ; i < count ; i++ )
            printf(" %d", num[i]+1);
        putchar(10);
    }
    return 0;
}

purplest 2011-08-17 00:44 发表评论

poj 2849 brainf*ck

purplest — Tue, 16 Aug 2011 16:01:00 GMT

brainf*ck，其实就是brainfuck，一种语言，http://zh.wikipedia.org/wiki/Brainfuck

模拟题

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

const int M=32768;
unsigned char str[M];
char in[200000], cmd[200000];
int n;

int main()
{
    scanf("%d%*c", &n);
    for ( int cases = 0 ; cases < n ; cases++ )
    {
        int t=0, i, j, point = 0;
        while ( gets(in), in[0] != 'e' )
        {
            for ( i = 0 ; in[i] ; i++ )
            {
                if ( ' ' == in[i] ) continue;
                else if ( '%' == in[i] ) break;
                else if ( '[' == in[i] ) t++;
                else if ( ']' == in[i] ) t--;
                cmd[point++]= in[i];
            }
        }
        cmd[point]=0;
        printf("PROGRAM #%d:\n", cases+1);
        if ( t != 0 )
        {
            printf("COMPILE ERROR\n");
            continue;
        }
        int p= 0;
        memset (str, 0, sizeof(str));
        for ( i = 0 ; cmd[i] ; i++ )
        {
            if ( cmd[i] == '>' )
                p=(p+1)%M;
            else if ( cmd[i] == '<' )
                p=(p+M-1)%M;
            else if ( cmd[i] ==  '+' )
                str[p]++;
            else if ( cmd[i] ==  '-' )
                str[p]--;
            else if ( cmd[i] ==  '.' )
                putchar(str[p]);
            else if ( cmd[i] ==  '[' && !str[p] )
            {
                i++;
                for ( t = 1; 0 != t ; i++ )
                {
                    if ( '[' == cmd[i] ) t++;
                    else if ( ']' == cmd[i] ) t--;
                }
                i--;
            }
            else if ( cmd[i] ==  ']' && str[p] )
            {
                i--;
                for ( t = -1; 0 != t ; i-- )
                {
                    if ( '[' == cmd[i] ) t++;
                    else if ( ']' == cmd[i] ) t--;
                }
                i++;
            }
        }
        putchar(10);
    }
    return 0;
}

purplest 2011-08-17 00:01 发表评论

poj 1049 Microprocessor Simulation

purplest — Fri, 12 Aug 2011 01:53:00 GMT

简单的模拟，读懂了题就会做了

#include <stdio.h>

char hax[17]="0123456789ABCDEF";
char ans[256];

int main()
{
    char a, b;
    while (1)
    {
        int i;
        for ( i = 0 ; i < 256 ; i++ )
        {
            if( EOF == scanf("%c", ans+i) ) return 0;
            ans[i]-=( ans[i] > '9' ? 55 : 48 );
        }
        getchar();
        if ( ans[0] == 8 ) break;
        int num = 0, flag= 1;
        char t;
        while (flag)
        {
            switch(ans[num++])
            {
                case 0:
                    a= ans[ (ans[num]<<4)+ans[num+1] ];
                    num+=2;
                    break;
                case 1:
                    ans[ (ans[num]<<4)+ans[num+1] ]=(a&=15);
                    num+=2;
                    break;
                case 2:
                    t= a;
                    a= b;
                    b= t;
                    break;
                case 3:
                    a+=b;
                    b=a>>4;
                    break;
                case 4:
                    a++;
                    break;
                case 5:
                    a--;
                    break;
                case 6:
                    if ( a&15 )
                    {
                        num+=2;
                        break;
                    }
                case 7:
                    num=(ans[num]<<4)+ans[num+1];
                    break;
                case 8:
                    flag=0;
                    break;
                default:
                    break;
            }
        }
        for ( i = 0 ; i < 256 ; i++ )
            printf("%c", hax[ans[i]]);
        putchar(10);
    }
    return 0;
}

purplest 2011-08-12 09:53 发表评论

poj 2846 The Bank of Kalii

purplest — Wed, 10 Aug 2011 16:36:00 GMT

历法类的题目，纯模拟，处理2月份的代码可能不是很精练，不过思路还是很清晰的

#include <stdio.h>

int judgeleap(int year)
{
    if ( ( !(year%4) && year%100 ) || !(year%400) )
        return 1;
    else return 0;
}

int leapmon[]={ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int mon[]={ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int n;

int main()
{
    scanf("%d", &n);
    for ( int l = 1; l <= n; l++ )
    {
        int year, day, month, inday, inmon, num, flag;
        scanf("%d/%d/%d%d/%d", &month, &day, &year, &inmon, &inday );
        printf("%d ", l);
        if ( month == inmon )
        {
            num= day-inday;
            num=num>0?num:-num;
            day>inday?(flag=0):(flag=1);
        }
        else if ( 1 == month && 12 == inmon )
        {
            num=31-inday+day;
            year--;
            flag= 0;
        }
        else if ( 1 == inmon && 12 == month )
        {
            num=31-day+inday;
            year++;
            flag= 1;
        }
        else if ( ( month < inmon-1 ) || ( inmon < month-1 ) )
        {
            num=8;
        }
        else if ( 2 == month )
        {
            if ( month < inmon )
            {
                if ( judgeleap(year) )
                {
                    num=29-day+inday;
                    flag= 1;
                }
                else
                {
                    num=28-day+inday;
                    flag= 1;
                }
            }
            else
            {
                num=31-inday+day;
                flag= 0;
            }
        }
        else if ( 2 == inmon )
        {
            if ( month < inmon )
            {
                num=31-day+inday;
                flag= 1;
            }
            else
            {
                if ( judgeleap(year) )
                {
                    num=29-inday+day;
                }
                else
                {
                    num=28-inday+day;
                }
                flag= 0;
            }
        }
        else if ( month < inmon )
        {
            if ( judgeleap(year) )
            {
                num=leapmon[month]-day+inday;
                flag= 1;
            }
            else
            {
                num=mon[month]-day+inday;
                flag= 0;
            }
        }
        else if ( inmon < month )
        {
            if ( judgeleap(year) )
            {
                num=leapmon[month]-inday+day;
                flag= 0;
            }
            else
            {
                num=mon[month]-inday+day;
                flag= 1;
            }
        }
        if ( 0 == num )
            printf("SAME DAY\n");
        else if ( 1 == num )
        {
            printf("%d/%d/%d IS %d DAY ", inmon, inday, year, num);
            if (flag) printf("AFTER\n");
            else printf("PRIOR\n");
        }
        else if ( 8 > num)
        {
            printf("%d/%d/%d IS %d DAYS ", inmon, inday, year, num);
            if (flag) printf("AFTER\n");
            else printf("PRIOR\n");
        }
        else
            printf("OUT OF RANGE\n");
    }
    return 0;
}

purplest 2011-08-11 00:36 发表评论

poj 1666 Candy Sharing Game

purplest — Wed, 10 Aug 2011 16:30:00 GMT

纯模拟，读懂题就能做，注意下hit就行了

#include <stdio.h>

int n;
int num[2000], t[2000];

int init()
{
    int i;
    for ( i = 0 ; i < n; i++ )
        if ( num[i]%2 )
            num[i]+=1;
    for ( i = 1; i < n ; i++)
        if ( num[0] != num[i] )
            return 1;
    return 0;
}

int main()
{
    while ( EOF != scanf("%d", &n) && n )
    {
        int i, m = 0;
        for ( i = 0; i < n ; i++ )
            scanf("%d", num+i);
        while (init())
        {
            m++;
            for ( i = 0; i < n; i++ )
            {
                num[i]>>=1;
                t[(i+1)%n]=num[i];
            }
            for ( i = 0 ; i < n ; i++ )
                num[i]+=t[i];
        }
        printf("%d %d\n", m, num[0]);
    }
    return 0;
}

purplest 2011-08-11 00:30 发表评论

poj 1051 P,MTHBGWB

purplest — Wed, 10 Aug 2011 16:28:00 GMT

一道摩尔电码翻译的题，用的模拟，还好数据量不大，如果数据量大了，得上map，或者字典树了。。

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

int n;
string chara[40]={"12", "2111", "2121", "211", "1", "1121", "221", "1111", "11", "1222", "212", "1211", "22", "21", "222", "1221",
"2212", "121", "111", "2", "112", "1112", "122", "2112", "2122", "2211", /*符号*/ "1122"/*_*/, "2221",/*.*/ "1212",/*,*/ "2222"/*?*/};

int main()
{
    while ( EOF != scanf("%d", &n) )
    {
        int q;
        for (q=1; q<=n; q++)
        {
            int num[100];
            char str[510];
            string ch="";
            scanf("%s", str);
            int i, j, k= 0, p= 0, len;
            for ( i = 0; str[i] != 0; i++ )
            {
                if ( 'A' <= str[i] && 'Z' >= str[i] )
                {
                    num[i]= chara[str[i]-'A'].length();
                    ch+=chara[str[i]-'A'];
                }
                else if ( str[i] == '_' )
                {
                    num[i]= chara[26].length();
                    ch+=chara[26];
                }
                else if ( str[i] == '.' )
                {
                    num[i]= chara[27].length();
                    ch+=chara[27];
                }
                else if ( str[i] == ',' )
                {
                    num[i]= chara[28].length();
                    ch+=chara[28];
                }
                else if ( str[i] == '?' )
                {
                    num[i]= chara[29].length();
                    ch+=chara[29];
                }
                k++;
            }
            printf("%d: ", q);
            for ( i= k-1; i >= 0; i--)
            {
                len= num[i];
                string t;
                for ( j= 0; j < len; j++)
                    t+=ch[j+p];
                for ( j = 0; j < 30 && t != chara[j]; j++);
                if ( 0 <= j && 26 > j) printf("%c", j+'A');
                else if ( 26 == j ) printf("_");
                else if ( 27 == j ) printf(".");
                else if ( 28 == j ) printf(",");
                else if ( 29 == j ) printf("?");
                p+=len;
            }
            putchar(10);
        }
    }
    return 0;
}

purplest 2011-08-11 00:28 发表评论

poj 1219 L-I-N-G-O: LINGO

purplest — Wed, 10 Aug 2011 16:22:00 GMT

纯模拟题，注意各种条件，比如开始的空白行要getchar()掉，放弃解答等

#include <cstdio>
#include <string>
#include <iostream>
#include <cstring>
using namespace std;

char in[100], str[100];

int upper(string t)
{
    for ( int i = 0; i < 5; i++ )
    {
        if ( 'A' > t[i] || 'Z' < t[i] )
            return 0;
    }
    return 1;
}
void change(string t, string & last)
{
    int i, j;
    int flag[10];
    memset(flag, 0, sizeof(flag));
    for ( i = 0; i < 5 ; i++ )
    {
        if (t[i] == in[i]) flag[i]=1;
    }
    for ( i = 0; i < 5 ; i++ )
    {
        if (t[i] == in[i]) last+=t[i];
        else
        {
            for ( j = 0 ; j < 5 ; j++ )
            {
                if(t[i] == in[j] && !flag[j])
                {
                    last+=in[j]-'A'+'a';
                    flag[j]=1;
                    break;
                }
            }
            if ( 5 == j ) last+= ".";
        }
    }
}

int main()
{
    getchar();
    while ( gets(in), 0 != strcmp(in, "LINGO") )
    {
        putchar(10);
        printf("%c....\n", in[0]);
        string t, gin=in, last;
        last=in[0];
        last+="....";
        int i, j, flag= 0;
        for (i= 0; i < 6; i++ )
        {
            gets(str);
            t=str;
            //printf("t=%s\n", t.c_str());
            if ( t == gin )
            {
                printf("%s\n", gin.c_str());
                break;
            }
            else if ( t == "" )
            {
                flag=1;
                break;
            }
            else
            {
                if(5 == t.length() && upper(t) )
                {
                    last="";
                    change(t, last);
                }
                if ( 5 == i )
                {
                    flag= 1;
                    break;
                }
                printf("%s\n", last.c_str());
            }
        }
        if ( flag )
        {
            for ( j = 0; j < 5; j++ )
                printf("%c", gin[j]-'A'+'a');
            putchar(10);
        }
        while( 0 != strcmp(str, "") )
        {
            gets(str);
        }
    }
    return 0;
}

purplest 2011-08-11 00:22 发表评论

poj 1665 Biker's Trip Odometer

purplest — Fri, 05 Aug 2011 15:14:00 GMT

简单的模拟

#include <stdio.h>

const float pi=3.1415927;

float d, time, dis, mph;
int rev;

int main()
{
    int n = 1;
    while ( scanf("%f%d%f", &d, &rev, &time) && rev )
    {
        dis=(d*pi*rev)/63360;
        mph=dis/time*3600;
        printf("Trip #%d: %.2f %.2f\n", n, dis, mph);
        n++;
    }
    return 0;
}

purplest 2011-08-05 23:14 发表评论

poj 2272 Bullseye

purplest — Fri, 05 Aug 2011 15:12:00 GMT

纯模拟

#include <stdio.h>

double dist(double x, double y)
{
    return x*x+y*y;
}

int main()
{
    int sum[2];
    double x[2][3], y[2][3];
    while ( scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf", &x[0][0], &y[0][0], &x[0][1], &y[0][1], &x[0][2], &y[0][2], &x[1][0], &y[1][0], &x[1][1], &y[1][1], &x[1][2], &y[1][2]), x[0][0] != -100.0 )
    {
        sum[0]=0, sum[1]=0;
        double dis;
        int i, j;
        for ( j = 0; j < 2; j++ )
        {
            for ( i = 0; i < 3; i++ )
            {
                dis= dist( x[j][i], y[j][i] );
                if (dis <= 9.0) sum[j]+=100;
                else if ( dis <= 36.0 ) sum[j]+=80;
                else if ( dis <= 81.0 ) sum[j]+=60;
                else if ( dis <= 144.0 ) sum[j]+=40;
                else if ( dis <= 225.0 ) sum[j]+=20;
            }
        }
        printf("SCORE: %d to %d, ", sum[0], sum[1]);
        if ( sum[0] > sum[1] ) printf("PLAYER 1 WINS.\n");
        else if ( sum[0] < sum[1] ) printf("PLAYER 2 WINS.\n");
        else printf("TIE.\n");
    }
    return 0;
}

purplest 2011-08-05 23:12 发表评论

poj 1068 Parencodings

purplest — Fri, 05 Aug 2011 15:00:00 GMT

纯模拟

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

int seq[1000];
int n;

int main()
{
    scanf("%d", &n);
    while (n--)
    {
        memset(seq, 0, sizeof(seq));
        int t, m, p=0, i, num= 1, j;
        scanf("%d", &t);
        while (t--)
        {
            scanf("%d", &m);
            for ( i = p; num <= m; i ++ )
            {
                seq[i]=1;
                num++;
            }
            seq[i]=2;
            p=i+1;
        }
        for ( i = 0; i < p; i++ )
        {
            if ( seq[i] == 2 )
            {
                num=1;
                for ( j = i-1; j >= 0 && seq[j] != 1; j-- )
                    if (seq[j]==3) num++;
                printf("%d ", num);
                seq[j]=3;
            }
        }
        putchar(10);
    }
    return 0;
}

purplest 2011-08-05 23:00 发表评论