Description：
给定一个多项式(by+ax)^k ，请求出多项式展开后x^n * y^m 项的系数
Analysis：
[C_k^m imes a^n imes b^m
]
Code
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mod = 10007;
typedef long long ll;
ll Pow(ll a,ll b)
{
ll c = 1;
for(;b;b >>= 1)
{
if(b&1)
{
c = a*c%mod;
}
a = a*a%mod;
}
return c%mod;
}
ll tri[1011][1011],a,b,k,n,m;
int main()
{
scanf("%d%d%d%d%d",&a,&b,&k,&n,&m);
tri[1][1] = 1;
for(int i = 2;i <= k + 1;++i)
{
for(int j = 1;j <= i;++j)
{
tri[i][j] = (tri[i-1][j] + tri[i-1][j-1])%mod;
}
}
ll ans = (tri[k + 1][m + 1]%mod * Pow(b,m)%mod * Pow(a,n)%mod)%mod;
printf("%lld",ans);
return 0;
}