hud 1061 Rightmost Digit

Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44105 Accepted Submission(s): 16599

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

Recommend

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1061
题目大意：
正整数n，求出n^n的最右边的数字。
解题思路：
每次用个位乘，这个就不说了，要乘n次，直接循环会超时，很容易想到递归思想，深搜即可。
代码如下：

#include <cstdio>
#include <cstring>
int n,x;

int dfs(int cnt)
{
    if(cnt==1)
        return x;
    else if(cnt%2==0)
    {
        int nx=dfs(cnt/2)*dfs(cnt/2);
        return nx%10;
    }
    else
    {
        int nx=dfs(cnt/2)*dfs(cnt/2)*x;
        return nx%10;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        x=n%10;
        printf("%d\n",dfs(n));
    }
    return 0;
}