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Rightmost Digit（HDU1061）

2013-08-16 09:04 155 查看

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 23184    Accepted Submission(s): 8836


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.

 
//解题思路：N(1<=N<=1,000,000,000).范围比较大，所以这题应该是有规律的，所以先打表找规律；
#include <cstdio>
#include <cstring>

int main()
{
int num[25] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
int i , j;
/*freopen("data.txt" ,"w" ,stdout);//打表找规律，发现每20个数的尾数都是相同的；用数组储存每一个数的尾数，然后直接输出
for(i = 1; i <= 100 ; i++)
{
int sum = 1;
for(j = 1 ; j <= i; j++)
{
sum *= i;
if(sum > 100)
sum %= 100;
}
printf("%d " ,sum%10);
if(i % 20 == 0)
printf("\n");
}*/
int n ;
while(scanf("%d" ,&n) != EOF)
{
int nums;
while(n--)
{
scanf("%d" ,&nums);
while(nums > 20)
{
nums %= 20;
}
printf("%d\n" , num[nums]);
}
}
}

                                            

                                        

                        
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