HDU--1061
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39389 Accepted Submission(s): 14863
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).
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 6HintIn 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
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
long long int powermod(long long int a,long long int b,long long int c)
{
    long long int ans = 1;
    a = a%c;
    while (b > 0)
    {
        if (b% 2 ==1)
            ans = ans * a %c;
        b = b/2;
        a = a* a %c;
    }
    return ans;
}
int main()
{
    int t;
    long long int n;
    cin >> t;
    while (t--)
    {
        cin >> n;
        cout <<powermod(n,n,10) << endl;
    }
    return 0;
}快速幂求余!
 
                     
                    
                 
                    
                
 
 
                
            
         
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浙公网安备 33010602011771号