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Problem Description
Given a positive integer N, you should output the most right digit of N^N.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).
Output
For each test case, you should output the rightmost digit of N^N.
Sample Input
2
3
4
Sample Output
7
6
快速幂：
#include<bits/stdc++.h>
using namespace std;
inline int read(){
int x=0,f=0;char ch=getchar();
while(ch>'9'||ch<'0')f|=ch=='-',ch=getchar();
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
return f?-x:x;
}
long long qpow(long long a,long long b){
long long ans=1;
while(b){
if(b&1)ans=ans*a%10;
a=a*a%10;
b>>=1;
}
return ans%10;
}
int main(){
int t=read(),p;
while(t--)p=read(),printf("%lld\n",qpow(p,p));
return 0;
}
打表：
#include<bits/stdc++.h>
using namespace std;
inline int read(){
int x=0,f=0;char ch=getchar();
while(ch>'9'||ch<'0')f|=ch=='-',ch=getchar();
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
return f?-x:x;
}
int f[20]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
int main(){
int t=read();
while(t--)printf("%d\n",f[read()%20]);
return 0;
}