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# Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19870    Accepted Submission(s): 7659

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.

Author
Ignatius.L

```#include<stdio.h>
#include<string.h>
int main()
{
int ans, n, m, i, j, t, circle[100];
scanf("%d", &m);
for(i=1; i<=m; ++i)
{
memset(circle, 0, sizeof(circle));
scanf("%d", &n);
circle[0] = n%10;
for(j=1; j<100; ++j)
{
circle[j] = (circle[j-1]*(n%10))%10;
if(circle[0] == circle[j]) break;
}

if(n%j == 0) printf("%d\n", circle[j-1]);
else printf("%d\n", circle[n%j - 1]);
}
return 0;
}```

解题报告：

各种水，天知道我上辈子欠了谁，2WA，原因是没有注意到溢出，可是这题做过啊！

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• 原文地址：https://www.cnblogs.com/liaoguifa/p/2745516.html