Rightmost Digit （求n^n最后一位）

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.

n^n最后一位，等价于(n%100)^(n%100)的最后一位。

 #include<cstdio>

 int main()

 {

     __int64 t,l,b;

     scanf("%I64d",&t);

     while(t--)

     {

         __int64 n;

         scanf("%I64d",&n);

             l=n%;

             b=l;

             for(__int64 i=;i<l;i++)

             {

                 b=b*l;

                 b%=;

             }

             b%=;

             printf("%I64d\n",b);

     }

 }

快速幂

快速幂求n^n;

 1 int f1(int a,int b)

 2 {

 3     int t=1;

 4     while(b)

 5     {

 6         if(b % 2 != 0)

 7         {

 8             t*=a;

 9             b--;

10         }

11         a*=a;

12         b/=2;

13     }

14     return t;

15 }

快速幂求n^n后y位；

 1 int f2(int a,int b)

 2 {

 3     int t=1;

 4     while(b)

 5     {

 6         if(b % 2 != 0)

 7         {

 8             t=(t*a)%x;    //x控制要求的位数

 9             b--;

10         }

11         a=(a*a)%x;

12         b/=2;

13     }

14     return t;

15 }

代码

 #include<cstdio>

 __int64 f(__int64 a)

 {

     __int64 b=a;

     __int64 t=;

     while(b)

     {

         if(b%!=)

         {

             t=(t*a)%;

             b--;

         }

         a=a*a%;

         b/=;

     }

     return t;

 }

 int main()

 {

     __int64 t,a;

     scanf("%I64d",&t);

     while(t--)

     {

         scanf("%I64d",&a);

         printf("%I64d\n",f(a));

     }

 }

巴特西