本文主要是介绍CSU - 1556 Jerry's trouble(快速幂取模),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
【题目链接】:click here
【题目大意】:计算x1^m+x2^m+..xn^m(1<=x1<=n)( 1 <= n < 1 000 000, 1 <= m < 1000)
【解题思路】:快速幂取模
代码:
solution one:
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod=(LL)1e9+7;
LL pow_mod(LL a,LL p,LL n)
{if(p==0) return 1;LL ans=pow_mod(a,p/2,n);ans=ans*ans%n;if(p&1) ans=ans*a%n;return ans;
}
int n,m;
int main()
{while(scanf("%d%d",&n,&m)!=EOF){LL s=0;for(int i=1; i<=n; i++)s=(s+pow_mod(i%mod,m,mod))%mod;printf("%lld\n",s);}return 0;
}
solution two:
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;typedef long long LL;
const LL mod=1e9+7;
LL pow_mod(LL a,LL b)
{LL res=a,ans=1;while(b){if(b&1) ans=(res*ans)%mod;res=res*res%mod;b>>=1;}return ans;
}
int main()
{LL n,m;while(~scanf("%lld %lld",&n,&m)){LL s=0;for(int i=1; i<=n; ++i)s+=pow_mod(i%mod,m)%mod;printf("%lld\n",s%mod);}return 0;
}
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