Rightmost Digit

浏览数：62 / 时间：2015年06月09日

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).

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.

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=1;
    a=a%c;
    if(a==0)
    return 0;
    while(b>0)
    {
        if(b%2==1)
        ans=ans*a%c;
        b/=2;
        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,10));
    }
    return 0;
}