I - Rightmost Digit

最新推荐文章于 2024-07-27 14:45:00 发布

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本文介绍了一个算法问题，即如何计算正整数N的N次方后的最右一位数字。通过输入一系列测试用例，算法可以快速计算出每个测试用例中N^N的最右一位数字。

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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.

代码：
#include<cstdio>
int main()
{long long int n;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
long long int sum=1%10,k=n%10;
while(n>0)
{
if(n%2==1)
sum=sum*k%10;
k=k*k%10;
n/=2;
}
printf("%d\n",sum);
}
return 0;
}