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;
}