Rightmost Digit

最新推荐文章于 2018-10-26 20:31:00 发布

a121566780

最新推荐文章于 2018-10-26 20:31:00 发布

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文章标签： integer output input each

本文链接：https://blog.youkuaiyun.com/a121566780/article/details/6611256

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本文介绍了一个算法，用于求解正整数N的N次方后的最右侧数字。通过分析不同个位数的幂次规律，使用switch-case结构实现了高效计算。

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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.

<code>

#include<stdio.h>
void main()
{ int i,N,j;
scanf("%d",&i);
while(i--)
{ scanf("%d",&N);
    switch(N%10)
{ case 0: j=0; break;
case 1: j=1; break;
case 2: switch(N%4)
   { case 0: j=6;break;
           case 2: j=4;break;
   } break;
case 3: switch(N%4)
     {case 0: j=1;break;
         case 1: j=3;break;
         case 2: j=9;break;
         case 3: j=7;break;
   } break;
case 4: j=6;break;
case 5: j=5;break;
    case 6: j=6;break;
case 7: switch(N%4)
     {
         case 1: j=7;break;
         case 3: j=3;break;
     }break;
case 8: switch(N%4)
     {
         case 2: j=4;break;
         case 0: j=6;break;
     }break;
case 9: j=9;break;
}
printf("%d\n",j);
}
}