标签：
Rightmost Digit
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 87 Accepted Submission(s) : 38
Font: Times New Roman | Verdana |
Georgia
Font Size: ← →
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
Sample Output
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.
Author
Ignatius.L
求n的n次方并输出最后一位，因为是次方所以数据会很大可以考虑用大数。但是大数很麻烦，于是找规律。通过列举数据会发现n的n方的最后一位每四次一循环。所以最多只要循环乘4次就可以得到答案。
AC代码：
#include <iostream>
using namespace std;
int main(){
long n,m,num,p;
cin>>num;
while(num--){
cin>>n;
m = p = n % 10;
if (n % 4 == 0) n = 4;
else n = n % 4;
while(--n) p = (p * m) % 10;
cout<<p<<endl;
}
return 0;
}
运行结果：
HDU Rightmost Digit
标签：
原文地址：http://blog.csdn.net/zp___waj/article/details/46041501
