hdu 4248排列问题

一看就可以用母函数做，不过好久没练DP了，所以还是用DP做了。用dp[i][j]表示前i种石头排出j个出来的种数，当考虑第i种石头石，枚举其使用的个数即可。WA了好几次，是整数相乘精度的问题，使用long long就过了。

/*

 * hdu1004/win.cpp

 * Created on: 2012-7-24

 * Author    : ben

 */

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

#include <stack>

#include <string>

#include <vector>

#include <deque>

#include <list>

#include <functional>

#include <numeric>

#include <cctype>

using namespace std;

typedef long long LL;

const int MAXN = 105;

const int MAXV =10005;

const int MOD = 1000000007;

int nums[MAXN];

int dp[MAXN][MAXV];

int c[MAXV][MAXN];



void InitCombineNum(){

    memset(c, 0, sizeof(c));

    for (int i = 0; i < MAXV; i++) {

        c[i][0] = 1;

    }

    c[1][1] = 1;

    for (int i = 2; i < MAXV; i++) {

        for (int j = 1; j < MAXN; j++) {

            if(j == i) {

                c[i][j] = 1;

                break;

            }

            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;

        }

    }

}



int main() {

#ifndef ONLINE_JUDGE

    freopen("data.in", "r", stdin);

#endif

    int N, V, T = 0;

    InitCombineNum();

    while(scanf("%d", &N) == 1) {

        V = 0;

        for(int i = 0; i < N; i++) {

            scanf("%d", &nums[i]);

            V += nums[i];

        }

        memset(dp, 0, sizeof(dp));

        for(int i = 0; i < N; i++) {

            dp[i][0] = 1;

        }

        for(int j = 0; j <= nums[0]; j++) {

            dp[0][j] = 1;

        }

        for(int i = 1; i < N; i++) {

            for(int j = 1; j <= V; j++) {

                LL temp = 0;

                for(int k = 0; k <= nums[i] && k <= j; k++) {

                    temp += (((LL)dp[i - 1][j - k]) * c[j][k]) % MOD;

                }

                dp[i][j] = temp % MOD;

            }

        }

        LL ans = 0;

        for(int j = 1; j <= V; j++) {

            ans += dp[N - 1][j];

        }

        printf("Case %d: %d\n", ++T, (int)(ans % MOD));

    }

    return 0;

}