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  • HDU 1060  Leftmost Digit

       Leftmost Digit

     

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 21744    Accepted Submission(s): 8408


     

    Problem Description

    Given a positive integer N, you should output the leftmost 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 leftmost digit of N^N.

     

    Sample Input

    
     

    2 3 4

     

    Sample Output

    
     
    2 2

    Hint

    In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.

    题解:   {color{Red} n^{n}=a	imes 10^{x}
ightarrow lg\, n^{n}=x+lg\, a
ightarrow a=10^{nlgn-x}=10^{nlgn-(int)nlgn)}}

    #include<iostream>
    #include<string.h>
    #include<algorithm>
    #include<cmath>
    #define ll long long
    using namespace std;
    int main()
    {
            ll T;
            double n;
            scanf("%lld",&T);
            while(T--){
                    scanf("%lf",&n);
                    cout<<(ll)pow(10,n*log10(n)-(ll)(n*log10(n)))<<endl;
            }
            return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/aeipyuan/p/10704475.html
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