連結:http://acm.hdu.edu.cn/showproblem.php?pid=6063
Problem Description
Input
There are several test cases, please keep reading until EOF.
There are exact 10000 cases.
For each test case, there are 2 numbers n,k.
Output
For each test case, output “Case #x: y”, which means the test case number and the answer.
Sample Input
10 10
Sample Output
Case #1: 999999937
莫比烏斯反演 , 他們 說國小奧賽都學過 ,但我 沒發處理那個階乘,
是以 就10^10 mod 1e9+7 發現 剛好等于答案 ,然後 就開始了 暴力之路 。
但要注意2點 , 階乘用 快速幂 别傻傻的for循環, 就是每次* 的前後 和開始都要取模
代碼:
#include<bits/stdc++.h>
#define mod 1000000007
//#define mod 1e9+7
using namespace std;
long long pow(long long a,long long b)
{
long long int ans = ,base = a;
while(b!=)
{
if(b&){
base%=(long long)mod;
ans%=(long long)mod;
ans *= base;
ans%=(long long)mod;
}
base%=(long long)mod;
base *= base;
base%=(long long)mod;
b>>=;
}
return ans;
}
int main()
{
long long int n,k;
int t=;
while(~scanf("%lld %lld",&n,&k)){
t++;
printf("Case #%d: ",t);
printf("%lld\n",pow(n,k));
}
return ;
}
順便附上 整數快速幂 講解:
http://blog.csdn.net/changjiale110/article/details/76467887