hd1061 Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39554    Accepted Submission(s): 14930

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

234

 

 

Sample Output

76
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.
 

嗯，题意Hint上就看懂了

//n^n,算一下会发现每个数的从1次方到n次方最后一位是有规律的//循环周期为4例如7,1次方为7,2次方为9,3次方为3,4次方为1,5次方又为7如此循环 //当n很大时发现n的次方最后一位只和n的最后一位有关 #include<cstdio>#include<cstring>int main(){int t,n,c;int s;scanf("%d",&t);while(t--){s=1;scanf("%d",&n);c=n%4;if(!c)c=4;n=n%10;for(int i=1;i<=c;++i){s*=n;s=s%10;}printf("%d\n",s);}return 0;}