组合计数递推求法
题目描述
给定一个多项式 (ax+by)k,请求出多项式展开后 xnym项的系数。
样例
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int a,b,k,n,m;
const int N = 1010,MOD = 10007;
int c[N][N];
int main()
{
cin>>a>>b>>k>>n>>m;
a %= MOD, b %= MOD;
for (int i = 0; i <= k; i ++ )
for (int j = 0; j <= i; j ++ )
if(!j)c[i][j] = 1;
else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j ]) % MOD;
int res = c[k][n];
for (int i = 0; i < n; i ++ ) res = res * a % MOD;
for (int i = 0; i < m; i ++ ) res = res * b % MOD;
cout << res << endl;
return 0;
}