Codeforces Round #309 (Div. 2)C

Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color i before drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.

Input
The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.

Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).

The total number of balls doesn’t exceed 1000.

Output
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.

Examples
input
3
2
2
1
output
3
input
4
1
2
3
4
output
1680
Note
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:

1 2 1 2 3
1 1 2 2 3
2 1 1 2 3

这个题主要是练了一下组合取模

#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<string>#include<map>using namespace std;long long n,tu[1001],mod= 1000000007,fac[9000];long long qpow(long long a, long long b){    long long ans = 1; a %= mod;    for (long long i = b; i; i >>= 1, a = a * a % mod)        if (i & 1)ans = ans * a % mod;    return ans;}long long C(long long n, long long m){    if (m > n || m < 0)return 0;    long long s1 = fac[n], s2 = fac[n - m] * fac[m] % mod;    return s1 * qpow(s2, mod - 2) % mod;}int main(){#define int long long    fac[0] = 1;    for (int a = 1;a <= 1000;a++)fac[a] = fac[a - 1] * a%mod;    cin >> n;    for (int a = 1;a <= n;a++)cin >> tu[a];    int zhonglei = 1, he = tu[1];    for (int a = 2;a <= n;a++)    {        he += tu[a];        zhonglei *= C(he - 1, tu[a] - 1)%mod;        zhonglei %= mod;    }    cout << zhonglei;}