Light OJ 1226 One Unit Machine (大组合数计算+DP)

解析：

设dp[i]为考虑前i个任务的方案数,s[i]为前i 个任务的总时间。

则dp[0] = 1，s[0] = a[0]。

状态转移方程： dp[i] = dp[i-1]*C(s[i-1]+a[i]-1,a[i]-1)。

[code]:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>

using namespace std;
typedef long long LL;
const LL MOD = 1e9+7;
const int M = 1e6+6;

int n,a[1005];
LL mul[M],dp[1005],s[1005];

LL mod_pow(LL a,LL b){
    LL res = 1;
    while(b){
        if(b&1) res = (res*a)%MOD;
        a = (a*a)%MOD;
        b >>= 1;
    }
    return res;
}

LL Comb(LL n,LL m){
    if(n < m) return 0;
    if(m==0||m==n) return 1;
    LL res = (mul[n]*mod_pow(mul[m]*mul[n-m]%MOD,MOD-2))%MOD;
    return res;
}

int main(){
    int i,j,cas,T;
    scanf("%d",&cas);
    mul[0] = 1;
    for(i = 1;i < M;i++) mul[i] = mul[i-1]*i%MOD;
    for(T = 1;T <= cas;T++){
        scanf("%d",&n);
        for(i = 0;i < n;i++) scanf("%d",&a[i]);
        printf("Case %d: ",T);
        dp[0] = 1;s[0] = a[0];
        for(i = 1;i < n;i++){
            dp[i] = (dp[i-1]*Comb(s[i-1]+a[i]-1,a[i]-1))%MOD;
            s[i] = s[i-1]+a[i];
        }
        printf("%lld\n",dp[n-1]);
    }

    return 0;
}