HDOJ1061 Rightmost Digit[简单数学题]

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19050    Accepted Submission(s): 7329

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

 

 

 

 

 

 

 

 

code:

#include<iostream>
using namespace std;

int table[10][10]={
    0,0,0,0,0,0,0,0,0,0,
    1,1,1,1,1,1,1,1,1,1,
    2,4,8,6,2,4,8,6,2,4,
    3,9,7,1,3,9,7,1,3,9,
    4,6,4,6,4,6,4,6,4,6,
    5,5,5,5,5,5,5,5,5,5,
    6,6,6,6,6,6,6,6,6,6,
    7,9,3,1,7,9,3,1,7,9,
    8,4,2,6,8,4,2,6,8,4,
    9,1,9,1,9,1,9,1,9,1
};

int cnt[10]={10,10,4,4,2,10,10,4,4,2};

int main()
{
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int temp=n;
        temp%=10;
        if(!temp)
        {
            printf("0\n");
            continue;
        }
        printf("%d\n",table[temp][(n-1)%cnt[temp]]);
    }
    return 0;
}