题目:
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
Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
题意:找到n^n的最后一个数字
方法一: 找规律 ans[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
方法二:用快速幂的方法,结果为快速幂之后对10取余的数
代码:
1.
`//
// main.cpp
// hdu1061
//
// Created by zhan_even on 2018/10/29.
// Copyright © 2018年 zhan_even. All rights reserved.
//
#include
using namespace std;
int main(int argc, const char * argv[]) {
int ans[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
int n,t;
while(cin >> t){
while (t–) {
cin >> n;
cout << ans[n%20] <<endl;
}
}
return 0;
}
`
2
//
// main.cpp
// hdu1061
//
// Created by zhan_even on 2018/10/28.
// Copyright © 2018年 zhan_even. All rights reserved.
//
#include <iostream>
using namespace std;
typedef long long ll;
ll powerMod(ll a, ll n, int m){
ll res = 1;
while (n) {
if (n&1) {
res *= a;
res %=m;
}
a *= a;
a %= m;
n >>= 1;
}
return res;
}
int main(){
int t,n;
cin>>t;
while(t--) {
cin >> n;
cout << powerMod(n, n, 10) << endl;
}
return 0;
}