（hnust 1586）Maximum Product

时间限制: 1 Sec 内存限制: 128 MB
提交: 197 解决: 58

题目描述
Given a positive integer N, split it into K non-negative integers. i.e. N = x1+x2+..+xk. Define p as the product of all integers in the set. i.e. p=x1*x2…*xk.

What’s the maximum possible value of p ? Since p may be very large, just print p%1000000007.

输入
Each line contains two integers N and K separated by space.1≤ N≤ 1000000000

, 1≤ K≤100。 Process to end of file.

输出
For each case, you should output maximum product mod 1000000007.

样例输入
3 3
4 3
样例输出
1
2

分析：余数与商的关系

#include<cstdio>
typedef long long LL;
const int mod=1e9+7;
LL pow_mod(LL a,LL b,LL mod)
{
    LL ans=1;
    a%=mod;
    while(b)
    {
        if(b&1) ans=(ans*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return ans;
}
int main()
{
    LL n,k;
    while(~scanf("%lld%lld",&n,&k))
    {
        LL m=n%k,tmp=n/k;
        LL t=tmp,ans=t,num=1;
        if(m)
        {
            t=1+tmp;
            num=m;
            ans=pow_mod(t,m,mod)%mod;
        }
        LL r=k-num;
        for(int i=0; i<r; i++)
            ans=ans*((n-num*t)/r)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}