hdu1061-N*N问题

1061

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.

规律 ：

个位数是它自己本身的数：0 1 5 6 9

2：2 4 8 6 //输入个位为2的数的次方永远是偶数，于是只存在输出数字为加粗的数字的情况

3：3 9 7 1

4：4 6

7：7 1

输入n保存个位：

n%10

（；´д｀）ゞ错误代码↓

#include<iostream>
#include<string>
using namespace std;
int main()
{
int t,n;
while(cin>>t)
{
while(t--)
{
cin>>n;
n = n%10;  //错误根源！直接将n覆盖
if(n == 0 || n == 1 || n == 5 || n == 6 || n == 9) cout<<n<<endl;
if(n == 2){
if(n%4 == 2 )cout<<'4'<<endl; //假设此时n=2,那么会一直有 n%4 = 2 ，不执行else
else cout<<'6'<<endl;
}
if(n == 3){
if(n%4 == 1)cout<<'3'<<endl;
else cout<<'7'<<endl;
}
if(n == 4)
cout<<'6'<<endl;
if(n == 7){
if(n%4 == 1)cout<<'7'<<endl;
else cout<<'3'<<endl;
}
if(n == 8){
if(n%4 == 2 )cout<<'4'<<endl;
else cout<<'6';
}

}
}
return 0;
}

正确代码↓

#include<iostream>
#include<string>
using namespace std;
int main()
{
int t,n,m;
while(cin>>t)
{
while(t--)
{
cin>>n;
m = n%10;  //得用两个变量区分余数和个位
if(m == 0 || m == 1 || m == 5 || m == 6 || m == 9) cout<<m<<endl;
if(m == 2){
if(n%4 == 2 )cout<<'4'<<endl;
else cout<<'6'<<endl;
}
if(m == 3){
if(n%4 == 1)cout<<'3'<<endl;
else cout<<'7'<<endl;
}
if(m == 4)
cout<<'6'<<endl;
if(m == 7){
if(n%4 == 1)cout<<'7'<<endl;
else cout<<'3'<<endl;
}
if(m == 8){
if(n%4 == 2 )cout<<'4'<<endl;
else cout<<'6'<<endl;
}

}
}
return 0;
}