Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35848 Accepted Submission(s): 13581
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).
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
3
4
Sample Output
7
6
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.
思路：一道简单的快速幂，刚开始看错了题目，一直没有思路。
若把a^b%c看做b个a相乘并分布求余，b太大显然会超时，快速幂的原理是：降低 b 的次数，把a变大（涉及求余，对a不影响）；
#include<cstdio>
int powermod(int a,int b,int c)
{
int ans=;
a=a%c;
if(a==)
return ;
while(b>)
{
if(b%==)
ans=ans*a%c;
b/=;
a=a*a%c;
}
return ans;
}
int main()
{
int T;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
printf("%d\n",powermod(n,n,));
}
return ;
}