hud 1061 Rightmost Digit

最新推荐文章于 2021-05-01 16:38:27 发布

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#poj #dfs #hdu

本文介绍了一个使用递归方法解决计算大数幂运算最右位数字的问题，通过不断循环和递归操作，实现高效计算。

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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

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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;
}