HDOJ 题目1061Rightmost Digit（数学）

最新推荐文章于 2021-05-01 16:38:27 发布

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最新推荐文章于 2021-05-01 16:38:27 发布

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Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32251 Accepted Submission(s): 12400

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

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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思路： http://blog.youkuaiyun.com/mq891116/article/details/7540481

这题必然不可能完全算出N^N后取最后一位，无论是storage和time开销都无法承受。网上看到有同学用的二分法，大概是用二分法计算次幂吧，也超时了。解法在两个方面做出了大量优化：

1. 注意到求的是个位，两个数乘积的个位只和这两个数的个位有关，于是可以完全抛弃个位之前的所有数字。

2. 对于N次幂，如果不加简化，仍然非常大。考虑到N个N相乘，各位数字肯定会有一个循环产生，尝试了下果然如此，详见下表：

N	1	2	3	4	5	...
0	0	0	0	0	0	...
1	1	1	1	1	0	...
2	2	4	8	6	2	...
3	3	9	7	1	3	...
4	4	6	4	6	4	...
5	5	5	5	5	5	...
6	6	6	6	6	6	...
7	7	9	3	1	7	...
8	8	4	2	6	8	...
9	9	1	9	1	9	...

可见0-9十个数字的公共循环周期是4，于是整个问题被简化为“求一位数的4次以内幂”，运算量大大减少。

ac代码

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main()
{
    int n;int s,a,b,num;
    scanf("%d",&n);
    while(n--)
    {
        
        scanf("%d",&num);
        a=num%10;
        b=num%4;
        if(b==0)
            b=4;
        printf("%d\n",(int)pow((double)a,b)%10);
    }
}