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  • Rightmost Digit

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 35848    Accepted Submission(s): 13581


    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.
     
     
    思路:一道简单的快速幂,刚开始看错了题目,一直没有思路。
    若把a^b%c看做b个a相乘并分布求余,b太大显然会超时,快速幂的原理是:降低 b 的次数,把a变大(涉及求余,对a不影响);
    #include<cstdio>
    int powermod(int a,int b,int c)
    {
        int ans=1;
        a=a%c;
        if(a==0)
        return 0;
        while(b>0)
        {
            if(b%2==1)
            ans=ans*a%c;
            b/=2;
            a=a*a%c;
        }
        return ans;
    }
    int main()
    {
        int T;
        int n;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&n);
            printf("%d
    ",powermod(n,n,10));
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/orchidzjl/p/4269477.html
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