HDU 1061 Rightmost Digit(快速幂取模)

                                           Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45296    Accepted Submission(s): 17031

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.

Author

Ignatius.L

题解：快速幂取模！！！

#include<iostream>
#include<cstdio>
int qmod(long long a,long long b,int c)
{
  int ans=1;
  a=a%c;
  while(b>0){
    if(b&1)
    ans=(ans*a)%c;
    b=b>>1;
    a=(a*a)%c;
  }
  return ans;
}
int main()
{
  int n;  long long t;
  scanf("%d",&n);
  while(n--)
  {
    scanf("%lld",&t);
    printf("%d\n",qmod(t,t,10));
  }
  
  return 0;
}

听说是ans的周期是20......(无语。。。)

#include<stdio.h>
int main()
{
  int T,n;
  int a[25]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
  
  scanf("%d",&T);
  
  while(T--)
  {
    scanf("%d",&n);
    printf("%d\n",a[n%20]);
    
  }
  return 0;
}