数学-组合数学

http://codeforces.com/contest/553/problem/A
这道题目难点在于找到规律，另外求组合数也是难点，不能用常规求法
求组合数的文章如下
http://blog.csdn.net/wty__/article/details/20048467

A. Kyoya and Colored Balls
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
**

代码

**：

#include
#include
#include
#include
#include
#define LL long long
using namespace std;
const int MOD=1000000007;
int a[1002];
map <int, LL> m;

//分解质因数
//k为1或-1
void fun(int n, int k)
{
    for (int i = 2; i <= sqrt(n * 1.0); i++)
    {
        while (n % i == 0)
        {
            n /= i;
            m[i] += k;
        }
    }
    if (n > 1)
    {
        m[n] += k;
    }
}

//大数快速幂取模
LL quick_pow(LL a, LL b)
{
    LL ret = 1;
    while (b)
    {
        if (b & 1)  //只有在b的二进制最后一位是1的时候才为正确
        {
            ret *= a;
            ret %= MOD;
        }
        b >>= 1;
        a *= a;
        a %= MOD;
    }
    return ret;
}

//求组合数
LL C(LL a, LL b)
{
    if (a < b || a < 0 || b < 0)
        return 0;
    m.clear();
    LL ret = 1;
    b = min(a - b, b);
    for (int i = 0; i < b; i++)
    {
        fun(a - i, 1);
    }
    for (int i = b; i >= 1; i--)
    {
        fun(i, -1);
    }

    ///以下计算出了具体的数
    for (__typeof(m.begin()) it = m.begin(); it != m.end(); it++)
    {
        if ((*it).second != 0)
        {
            ret *= quick_pow((*it).first, (*it).second);
            ret %= MOD;
        }
    }
    return ret;
}
int main()
{
    int n=0,k;
    while(cin>>k){
        for(int i=1;i<=k;i++){
            scanf("%d",&a[i]);
            n+=a[i];
        }
        long long  t=1,s=1;
        for(int i=k;i>=1;i--){
        t=t*C(n-1,a[i]-1)%MOD;
        n=n-a[i];
        }
        cout<return 0;

}