hdu 1061 Rightmost Digit

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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31559 Accepted Submission(s): 12070

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.

/*题解：
求N^N的最低位有一定规律。当N最末尾为0 1 5 6时，最末尾一定为0 1 5 6
当N最末尾为 2 3 4 7 8 9时，其次方的最末尾具有周期性。
*/

#include<cstdio>
int main(){
int T,t;
scanf("%d",&T);
while(T--){
__int64 m;
scanf("%I64d",&m);
t=m%10;
if(t==0) printf("0\n");
if(t==1) printf("1\n");
if(t==5) printf("5\n");
if(t==6) printf("6\n");
if(t==2){
if(m%4==0) printf("6\n");
if(m%4==1) printf("2\n");
if(m%4==2) printf("4\n");
if(m%4==3) printf("8\n");
}
if(t==3){
if(m%4==0) printf("1\n");
if(m%4==1) printf("3\n");
if(m%4==2) printf("9\n");
if(m%4==3) printf("7\n");
}
if(t==4){
if(m%2==0) printf("6\n");
if(m%2==1) printf("4\n");
}
if(t==7){
if(m%4==0) printf("1\n");
if(m%4==1) printf("7\n");
if(m%4==2) printf("9\n");
if(m%4==3) printf("3\n");
}
if(t==8){
if(m%4==0) printf("6\n");
if(m%4==1) printf("8\n");
if(m%4==2) printf("4\n");
if(m%4==3) printf("2\n");
}
if(t==9){
if(m%2==0) printf("1\n");
if(m%2==1) printf("9\n");
}
}
return 0;
}

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