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  • hdu1061 Rightmost Digit 快速幂的简单应用

    **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.
    **

    题意:求n的n次方最右边的数,也就是最后一个数字。
    思路:由于数值n的范围很大,不能用寻常套路写这道题,这道题的正解是快速幂取模运算的应用,基于公式:
    (a*b)%c=(a%c*b%c)%c
    (话说第一次WA就是因为取模出错,导致范围超了)

    #include<stdio.h>
    int f(int m,int n)
    {
        if( n == 1)
            return m;
        long long s = f(m,n/2)%10;
        if( n%2 == 1)
            return (s*s*m)%10;
        else
            return (s*s)%10;
    }
    
    int main()
    {
        int n;
        int t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            printf("%d
    ",f(n,n));
        }
        return 0;
     } 
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  • 原文地址:https://www.cnblogs.com/hellocheng/p/7350154.html
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