C++博客-Dreams

显示本地服务器时间代码

DreamSky — Mon, 31 Aug 2009 00:23:00 GMT

DreamSky 2009-08-31 08:23 发表评论

使用CSS控制页面的几种方法

DreamSky — Mon, 31 Aug 2009 00:22:00 GMT

1.行内样式

例如：

CSS内容

2.内嵌式

例如：

CSS内容

3.链接式

例如：

4.导入样式

例如：

DreamSky 2009-08-31 08:22 发表评论

TextBox and HtmlInputText

DreamSky — Wed, 12 Aug 2009 03:30:00 GMT

    实现一个小小的功能：获取焦点，费了很大力气
    控件代码：

      js函数：

nodes对象中存有两类控件：TextBox 和ImageButton，根据先后顺序，nodes[0]即表示TextBox控件，故可用如下方法获取焦点。

    document.getElementById("idName")可以获取前台的ID名为idName的控件。
    TextBox控件的 TextMode="MultiLine"可以实现自动断行，
    HtmlInputText控件还没研究出自动断行的方法，继续努力……

DreamSky 2009-08-12 11:30 发表评论

TOJ 1721 Friends

DreamSky — Mon, 18 May 2009 10:02:00 GMT

http://acm.tju.edu.cn/toj/showp1721.html

//640827 2009-05-18 17:58:06 Accepted 1721 C++ 1.6K 0'00.00" 1212K asliyunsong
#include<iostream>
#include<string>
#include<map>
using namespace std;

int main()
{
    int tc,T;
    int j,len;
    string str,names;
    string pep[] = {"a","Anne","Bob","Karin","Charly","Dave","Edward","Frank"};
    scanf("%d",&T);
    getchar();
    for(tc = 1; tc <= T;tc ++)
    {
        map<string,int> M;
        getline(cin,str);
        str += " ";
        len = str.length();
        names = "";
        for(j = 0;j<len;j++)
        {
            if(str[j] == ' ')
            {
                M[names] = 1;
                names = "";
            }
            else
                names += str[j];
        }
        int pt1 ,pt2,pt3;
        pt1 = pt2 = pt3 = 0;
        if(M[pep[1]] == 1)
            pt1++;
        if(M[pep[2]] == 1)
        {
            if(M[pep[3]] == 1 && M[pep[5]] == 0 && M[pep[1]] == 1 && M[pep[6]] == 0)
                pt2++;
            else if(M[pep[3]] == 0 && ( M[pep[5]] == 1 || M[pep[1]] == 0 || M[pep[6]] == 1))
                pt3 ++;
        }
        if(M[pep[3]] == 1)
        {
            if(M[pep[4]] == 1 )
                pt2++;
            else if(M[pep[4]] == 0 && M[pep[1]] == 1 )
                pt1 ++;
            else if(M[pep[4]] == 0 && M[pep[1]] == 0 )
                pt3 ++;
        }
        if(M[pep[4]] == 1)
        {
            if(M[pep[1]] == 1 )
                pt1++;
        }

        if(M[pep[6]] == 1)
        {
            if(M[pep[4]] == 0 && M[pep[1]] == 1 )
                pt3++;
            else
                pt1 ++;
        }

        if(M[pep[7]] == 1)
        {
            if(M[pep[2]] == 0 && M[pep[1]] == 0 )
                pt1 ++;
            else if(M[pep[1]] == 1 )
                pt2 ++;
        }

        printf("Scenario #%d:\n",tc);

        if(pt1 > pt2 && pt1 > pt3)
        {
            cout<<"cinema"<<endl;
        }
        else if(pt2 > pt1 && pt2 > pt3)
        {
            cout<<"disco"<<endl;
        }
        else if(pt3 > pt2 && pt3 > pt1)
        {
            cout<<"cocktail bar"<<endl;
        }
        else
            cout<<"stay at the Hacienda"<<endl;

        cout<<endl;
    }
    return 0;
}

DreamSky 2009-05-18 18:02 发表评论

hdu 2372 El Dorado

DreamSky — Thu, 14 May 2009 11:49:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=2372

//1367544 2009-05-14 19:45:52 Accepted 2372 0MS 352K 678 B C++ no way
#include<iostream>
using namespace std;
const int N = 105;
__int64 dp[N][N];//dp[i][j]表示一第i个元素结尾的长度为j的单调序列个数
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        if(n == 0 && k == 0)
            break;
        int i,j,p;
        __int64 num[N],temp;
        for(i=1;i<=n;i++)
            scanf("%I64d",&num[i]);

        for(j=2;j<=k;j++)
            for(i=j;i<=n;i++)
                dp[i][j] = 0;

        for(i=1;i<=n;i++)
            dp[i][1] = 1;

        for(j=2;j<=k;j++)
            for(i=j;i<=n;i++)
            {
                for(p=j-1;p<i;p++)
                    if(num[i] > num[p])
                        dp[i][j] += dp[p][j-1];
            }

        for(temp=0,i=k;i<=n;i++)
            temp += dp[i][k];

        printf("%I64d\n",temp);
    }
    return 0;
}

DreamSky 2009-05-14 19:49 发表评论

zju 2750 Idiomatic Phrases Game

DreamSky — Thu, 14 May 2009 03:09:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2750

//1868088 2009-05-14 08:38:13 Wrong Answer  2750 C++ 130 4100 aslys
//1868170 2009-05-14 09:46:00 Time Limit Exceeded  2750 C++ 1001 4100 aslys
//1868224 2009-05-14 10:24:24 Accepted  2750 C++ 150 4112 aslys
#include<iostream>
#include<string>
#include<queue>
using namespace std;

const int inf = 1000000000;
const int N = 1001;
struct Node
{
    int t;
    char s[5];
    char e[5];
}pt[N];
struct node
{
    int v;
    int cost;
    node(){}//定义中间变量时，要用到该析构函数，此时不需要初值
    node(int a,int c)
    {
        v = a;
        cost = c;
    }
    friend bool operator <(node a,node b) //重载运算符，从小到大排序
    {
        return a.cost > b.cost;
    }
};
int n,gra[N][N];

void dijkstra(int u)
{
    priority_queue<node>Q;
    node p,q;
    bool mark[N];
    int dist[N];
    int i,j,index;
    for(i=1;i<=n;i++)
    {
        if(gra[u][i] != inf)
        {
            p.v = i;
            p.cost = gra[u][i];
            Q.push(p);
        }
        dist[i] = gra[u][i];
        mark[i] = 0;
    }
    mark[u] = 1;
    for(i=1;i<=n;i++)
    {
        index = -1;
        while(!Q.empty()) //对头元素就是最小的
        {
            q = Q.top();
            Q.pop();
            if(mark[q.v] == 0)
            {
                index = q.v;
                break ;
            }
        }
        if(index == -1)
            break;
        if(index == n)
            break;
        mark[index] = 1;

        for(j=1;j<=n;j++)
            if(mark[j] == 0 && dist[j] > q.cost + gra[index][j])
            {
                dist[j] = q.cost + gra[index][j];
                p.cost = dist[j];
                p.v = j;
                Q.push(p);
            }
    }
    if(dist[n] == inf)
        printf("-1\n");
    else
        printf("%d\n",dist[n]);
}
int main()
{

    while(cin>>n && n)
    {
        char temp[N];
        int i,j,len;
        for(i=1;i<=n;i++)//init()
            for(j=1;j<=n;j++)
                gra[i][j] = inf;

        for(i=1;i<=n;i++)
        {
            scanf("%d%s",&pt[i].t,temp);

            len = strlen(temp);

            for(j=0;j<4;j++)//取头
                pt[i].s[j] = temp[j];
            pt[i].s[j] = '\0';

            len -= 4;
            for(j=0;j<4;j++)//取尾
                pt[i].e[j] = temp[j+len];
            pt[i].e[j] = '\0';
        }
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                if(strcmp(pt[i].e,pt[j].s) == 0)
                    gra[i][j] = pt[i].t;
        dijkstra(1);
    }
    return 0;
}

DreamSky 2009-05-14 11:09 发表评论

hdu 2437 Jerboas

DreamSky — Wed, 13 May 2009 12:17:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=2437

//1364284 2009-05-13 20:11:33 Accepted 2437 187MS 4444K 1362 B C++ no way
#include <iostream>
#include <vector>
using namespace std;

const int N = 1004;

typedef struct
{
    int vex;
    int cost;
}node;

vector<node> vv[N];

int ans, e;
char path[N];
int num[N][N];//num[i][j]表示走到i结点花费mod K 的最小值
int n, m, s, k;
int pcnt;

void init()
{
    for(int i = 1; i <= n; i++)
        vv[i].clear();
    ans = e = -1;
    memset(num, -1, sizeof(num));
}

void make_graph()
{
    int a, b, cost;
    node cc;

    for(int i = 0; i < m; i++)
    {
        scanf("%d%d%d", &a, &b, &cost);

        cc.vex = b; cc.cost = cost;
        vv[a].push_back(cc);
       // follows the tunnels you can not go back to the starting burrow.
        //表示单向图
       // cc.vex = a;
        //vv[b].push_back(cc);
    }
}

void dfs(int v,int costs)
{
    int i,j,p,t;
    if(path[v] =='P' && costs % k == 0 && (ans ==-1 || ans > costs || (ans == costs && v < e)))
    {
        ans = costs;
        e = v;
    }
    j = vv[v].size();
    if(!j)
        return ;
    for(i=0;i<j;i++)
    {
        p = vv[v][i].vex;
        t = vv[v][i].cost + costs;
        if(num[p][t%k] == -1 || num[p][t%k] > t)
        {
            num[p][t%k] = t;
            dfs(p,t);
        }
    }
}

int main()
{
    int T,i, kk;
    scanf("%d", &T);
    for(i = 1; i <= T; i++)
    {
        scanf("%d%d%d%d", &n, &m, &s, &k);
        getchar();
        scanf("%s", path + 1);
        init();
        make_graph();
        dfs(s,0 );
        cout<<"Case "<<i<<": "<<ans<<" "<<e<<endl;
    }
    return 0;
}

DreamSky 2009-05-13 20:17 发表评论

zju 2853 Evolution

DreamSky — Tue, 12 May 2009 01:55:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2853

//1865331 2009-05-11 21:26:46 Accepted  2853 C++ 3790 1484 aslys
#include<iostream>
using namespace std;
const int N = 204;
int n;
double matrix[N][N];
double x[N][N],y[N][N],z[N][N];
void fun(int k,double m[][N])
{
    int i,j,p,t;
    double sum;
    if(k == 1)  //矩阵的一次方
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                m[i][j] = matrix[i][j];
    }
    else if(k == 2) //矩阵的二次方
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            {
                sum = matrix[i][0] * matrix[0][j];
                for(p=1;p<n;p++)
                    sum += matrix[i][p] * matrix[p][j];
                m[i][j] = sum;
            }
    }
    else  //矩阵的 k 次方，先计算出 k/2 次方，加快计算速度
    {
        t = k/2;
        fun(t,x); //递归调用
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            {
                sum = x[i][0] * x[0][j];
                for(p=1;p<n;p++)
                    sum += x[i][p] * x[p][j];
                y[i][j] = sum;
            }
        if( k % 2 == 1)      // 比如对矩阵A ：A^9 = A^4 * A^4 * A
        {
            for(i=0;i<n;i++)
                for(j=0;j<n;j++)
                {
                    sum = matrix[i][0] * y[0][j];
                     for(p=1;p<n;p++)
                         sum += matrix[i][p] * y[p][j];;
                    m[i][j] = sum;
            }
        }
        else
        {
            for(i=0;i<n;i++)
                for(j=0;j<n;j++)
                    m[i][j] = y[i][j];
        }
    }
}

int main()
{
    int m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if( m==0 && n==0 )
            break;
        int i,j,k;
        double num[N],r;
        for(i=0;i<n;i++)//init()
            for(j=0;j<n;j++)
            {
                if(i == j)
                    matrix[i][j] = 1.0;
                else
                    matrix[i][j] = 0.0;
            }

        for(i=0;i<n;i++)
            scanf("%lf",&num[i]);

        scanf("%d",&k);
        while(k--)
        {
            scanf("%d%d%lf",&i,&j,&r);
            matrix[j][i] += r; //构建初始矩阵
            matrix[i][i] -= r;
        }

        fun(m,z); //求矩阵的m次方，结果放在 x 数组中

        r = 0.0;
        for(j=0;j<n;j++)
            r += z[n-1][j] * num[j];
        printf("%.0lf\n",r);
    }
    return 0;
}
/*
3 1
40 20 10
2
0 1 1
1 2 1
3 2
40 20 10
2
0 1 1
1 2 1
*/

DreamSky 2009-05-12 09:55 发表评论

zju 2974 Just Pour the Water

DreamSky — Mon, 11 May 2009 11:33:00 GMT

摘要: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1973 //1864877 2009-05-11 19:06:51 Accepted 2974 C++ 0 188 aslys //烦了点，却也无心改进了~ &nbs... 阅读全文

DreamSky 2009-05-11 19:33 发表评论

01-package

DreamSky — Fri, 08 May 2009 13:48:00 GMT

http://info.zjfc.edu.cn/acm/contest/contest_problemDetail.aspx?pid=1005&cid=32
题目描述：
    给定一个背包的容量k，给定n个物品的体积和价值，物品不可分割，将n个物品中选若干个物品放入背包，求背包内物品的最大价值总和，在价值总和最大的前提下求背包内的最小物品个数c。
输出描述：
    第一行是一个整数t，表示测试数据的组数t。
   对于每组测试数据，第一行是两个整数n和k,表示物品的个数和背包的容量；
    接下来n行，每行两个整数，分别是物品的价值和体积。
输出描述：
输出背包内物品的最大价值v，在价值最大的前提下求背包内的最小物品个数c，中间用一个空格隔开。
样例输入：
1
3 10
4 5
6 5
10 10
样例输出：
10 1
作者：
xiewenxiu

//16829 2009-05-08 19:58:40 1005  Accepted 765MS 464K Visual C++ liyunsong
#include<iostream>
using namespace std;
struct Node
{
    int ns;//最小物品数
    int vs;//最大价值
}dp[2001];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int v[2001],w[2001];
        int n,c;
        int i,j;
        cin>>n>>c;
        for(i=1;i<=n;i++)
            scanf("%d%d",&v[i],&w[i]);
        for(j=0 ; j < w[1];j++)
        {
            dp[j].vs = 0;
            dp[j].ns = 0;
        }
        for(; j <= c;j++)
        {
            dp[j].vs = v[1];
            dp[j].ns = 1;
        }
        for(i=2;i<=n;i++)
        {
            for(j=c;j>=w[i];j--)
            {
                if(dp[j].vs < dp[j-w[i]].vs + v[i])
                {
                    dp[j].vs = dp[j-w[i]].vs + v[i];
                    dp[j].ns = dp[j-w[i]].ns + 1;
                }
                else if(dp[j].vs == dp[j-w[i]].vs + v[i] && dp[j].ns > dp[j-w[i]].ns + 1)
                    dp[j].ns = dp[j-w[i]].ns + 1;
            }
        }
        cout<<dp[c].vs<<" "<<dp[c].ns<<endl;
    }
    return 0;
}

DreamSky 2009-05-08 21:48 发表评论

zju 1883 Tight Words

DreamSky — Fri, 08 May 2009 07:24:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=883

#include<stdio.h>
#include<math.h>
int main()
{
    int k,n;
    while(scanf("%d%d",&k,&n)!=EOF)
    {
        int i,t;
        double f[11][101];
        for(i=0;i<=k;i++)
            f[i][1] = 1;
        for(t = 2; t<=n;t++)
        {
            for(i=0;i<=k;i++)
            {
                f[i][t] = f[i][t-1] ;
                if(i+1<=k)
                   f[i][t] += f[i+1][t-1];
                if(i-1>=0)
                   f[i][t] += f[i-1][t-1];
            }
        }
        double sum=0.0,p;
        for(i=0;i<=k;i++)
            sum += f[i][n];
        p = sum / pow((double)(k+1),(double)n);
        printf("%.5lf\n",p*100);
    }
    return 0;
}

DreamSky 2009-05-08 15:24 发表评论

zju 3201 Tree of Tree

DreamSky — Thu, 07 May 2009 06:36:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3326

//1858391 2009-05-07 14:27:05 Accepted  3201 C++ 0 228 aslys
#include<iostream>
#include<vector>
using namespace std;
const int N = 105;
int n,k;
int f[N][N],w[N];
vector<int>sum[N];
//学习使用vector
void dp(int x,int y)
{
    int i,j,temp[N];
    memset(temp,0,sizeof(temp));
    for(i=1;i<=k;i++)
    {
        for(j=1;j<=i;j++)
            if(f[x][j] + f[y][i-j] > temp[i])//合并出最大的
                temp[i] = f[x][j] + f[y][i-j];
    }
    for(i=1;i<=k;i++)
        if(f[x][i] < temp[i])
            f[x][i] = temp[i];
}

void dfs(int now,int pre)
{
    int len = sum[now].size();
    int i,cnt = 0;
    for(i=0;i<len;i++)
    {
        if(sum[now][i] != pre)
        {
            dfs(sum[now][i],now);
            f[now][1] = w[now];
            dp(now,sum[now][i]);
            cnt ++;
        }
    }
    if(cnt == 0)
        f[now][1] = w[now];
}

int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        int i,a,b,maxs;
        for(i=0;i<n;i++)
        {
            scanf("%d",&w[i]);
            sum[i].clear();
        }
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            sum[a].push_back(b);
            sum[b].push_back(a);
        }
        maxs = 0;
        memset(f,0,sizeof(f));
        dfs(0,-1);//从任一点出发都行
        for(i=0;i<n;i++)
            if(maxs < f[i][k])
                maxs = f[i][k];
            printf("%d\n",maxs);
    }
    return 0;
}

DreamSky 2009-05-07 14:36 发表评论

zju 3175 Number of Containers

DreamSky — Thu, 07 May 2009 04:07:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3216

//1858220 2009-05-07 11:33:22 Accepted  3175 C++ 70 184 aslys
// x*y = n ……
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int i;
        long long n,s,sum=0;
        scanf("%lld",&n);
        s = sqrt((double)n);
        for(i = 1;i<=s;i++)
            sum += n/i;
        sum *= 2;
        sum -= s*s;
        sum -= n;
        printf("%lld\n",sum);
    }
    return 0;
}

DreamSky 2009-05-07 12:07 发表评论

zju 2852 Deck of Cards

DreamSky — Wed, 06 May 2009 00:48:00 GMT

摘要: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1852 #includeusing namespace std;int dp[201][22][22][22];//dp[k][r][s][t]表示前k张card里面三堆分别为r点、s点、t点时的最优值int&nbs... 阅读全文

DreamSky 2009-05-06 08:48 发表评论

hdu 1267 下沙的沙子有几粒？

DreamSky — Tue, 05 May 2009 04:53:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1267

#include<iostream>
using namespace std;
int main()
{
    int i,j,len=1,r=0,temp=0;
    int a[201][400]={0};
    a[1][0]=1;
    for(i=2;i<=200;i++)//求1……200的阶乘
    {
        for(j=0;j<len;j++)
            a[i][j] = a[i-1][j] * i;
        for(j=0,r=0;j<len;j++)
        {
            temp = a[i][j] + r;
            a[i][j] = temp % 10;
            r = temp /10;
        }
        while(r)
        {
            a[i][len++] = r%10;
            r /= 10;
        }
    }
    int m,n;
    while(cin>>m>>n)
    {
        r = 0;
        len = 400;//改成300就错了
        int b[500]={0};
        for(j=0;j<len;j++)
        {
            temp = a[m+n][j]*(m+1-n) + r;
            b[j] = temp % 10;
            r = temp /10;
        }
        while(r)
        {
            b[len++] = r%10;
            r /= 10;
        }

        for(j=len-1,r=0;j>=0;j--)
        {            //除法从高位到低位
            temp=r*10+b[j];
            b[j]=temp/(m+1);
            r=temp%(m+1);
        }
        while(!b[len])        //处理高位的零位
            len--;

        for(i=m;i>1;i--)//除m!
        {
            for(j=len,r=0;j>=0;j--)
                {            //除法从高位到低位
                temp=r*10+b[j];
                b[j]=temp/i;
                r=temp%i;
                }
            while(!b[len])        //处理高位的零位
            len--;
        }

        for(i=n;i>1;i--)//除n!
        {
            for(j=len,r=0;j>=0;j--)
                {            //除法从高位到低位
                temp=r*10+b[j];
                b[j]=temp/i;
                r=temp%i;
                }
            while(!b[len])        //处理高位的零位
            len--;
        }

        for(j=len;j>=0;j--)
            cout<<b[j];
        cout<<endl;
    }
    return 0;
}

DreamSky 2009-05-05 12:53 发表评论