2007: sum of power
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 62 Solved: 28
[Submit][Status][Web Board]
Description
Calculate 
mod (1000000000+7) for given n，m.
Input
Input contains two integers n,m(1≤n≤1000,0≤m≤10).
Output
Output the answer in a single line.
Sample Input
10 0
Sample Output
10
【分析】取余运算，（a*b）%mod=(a%mod*b%mod)%mod;
不定义函数：
#include<bits/stdc++.h>
using namespace std;
const int maxn=1000000007;
int main()
{
int n,m;
long long sum=0,cnt=1;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
cnt=(i%maxn*cnt)%maxn;//,cout<<cnt<<endl;
// cout<<"i^m="<<cnt<<endl;
sum+=cnt,sum=sum%maxn;cnt=1;
}
cout<<sum<<endl;
return 0;
}定义函数：
#include<bits/stdc++.h>
using namespace std;
const int maxn=1000000007;
long long pow(int a,int b)
{
long long ans=1;
for(int i=0;i<b;i++)
{
ans*=a;
ans=ans%maxn;
}
return ans;
}
int main()
{
int n,m;
long long sum=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
sum+=pow(i,m);
sum=sum%maxn;
}
cout<<sum<<endl;
return 0;
}