HDOJ---1061 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


[align=left]Problem Description[/align]
Given a positive integer N, you should output the most right digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the rightmost digit of N^N.

[align=left]Sample Input[/align]

2 3 4

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
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;
}