Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6515    Accepted Submission(s): 2454

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

//Right most Digital//拿到此题第一感觉可以通过两个数相乘然后求得最后一位，但是发现其实结果只和最后一位//有关，用整个数来求得最后一位未免太过麻烦，便考虑去寻找规律//便是 0所有相乘都为0,1所有相乘都为1，2相乘便是48624862...然后每一个都是循环关系，//循环周期分别为1,1,4,4,2,1,1,4,4,2//然后问题便转化为求末位，找规律，得结果 #include<iostream>#include<string>using namespace std;#define MAX 10000long int n,m;int main(){int t,u;    int a3[4]={1,3,9,7},a4[2]={6,4};    int a2[4]={6,2,4,8},a7[4]={1,7,9,3};    int a8[4]={6,8,4,2},a9[2]={1,9}; int x[10]={1,1,4,4,2,1,1,4,4,2};//0-9分别循环次数为此数组对应数位 while(cin>>t){while(t--){cin>>n;//幂次，n的n次方while(n!=0){m=n%10;n=n/10;} //除10取余，获取最后一位数字u=m; u=u%x[m];//u即为所除余数 if(m==0||m==1||m==5||m==6){cout<<m<<endl;} else if(m==2)      cout<<a2[u]<<endl;else if(m==3)      cout<<a3[u]<<endl;else if(m==4)      cout<<a4[u]<<endl;else if(m==7)      cout<<a7[u]<<endl;else if(m==8)      cout<<a8[u]<<endl;else if(m==9)      cout<<a9[u]<<endl;}}}