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  • 快速幂 HDU 1061 Rightmost Digit *

    Rightmost Digit

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 57430    Accepted Submission(s): 21736


    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

    快速幂的运用,每次相乘时余十就行

    注意碰到次方很多的时候可以考虑使用快速幂了

    这里贴一份讲快速幂的博客  http://blog.csdn.net/hikean/article/details/9749391

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #define ll long long
    #define mod 1000000007
    using namespace std;
    int main()
    {
        int T;
        cin >> T;
        while(T--){
            int n;
            cin >> n;
            int ans = 1,mul = n,num = n;
            mul = mul % 10;
            while(num){
                if(num%2==1){
                    ans = (ans*mul)%10;
                }
                mul = (mul*mul)%10;
                num /= 2;
            }
            cout << ans%10 << endl;
        }
        return 0;
    }
    彼时当年少,莫负好时光。
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  • 原文地址:https://www.cnblogs.com/l609929321/p/7272847.html
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