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  • N^N hdu1060

    Leftmost Digit
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8784    Accepted Submission(s): 3391
    
    
    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.
     
     
    
    Author
    Ignatius.L
    
    
    
    
    #include<iostream>
    #include<stdlib.h>
    #include<stdio.h>
    #include<math.h>
    using namespace std;
    int powN(int N)
    {
        int n;
        double temp,k;
        temp=log10((double)N);
        n=temp;
        k=temp-n;
        temp=N*k;
        n=temp;
        k=temp-n;
        /*开始的时候直接把N*1og(N),结果double超出了范围,
        后来我就先把temp变成小数再乘,精髓是取lg哦亲,元方你怎么看*/
        n=pow(10.0,k);
        return n;
    }
    int main()
    {
        int T,N;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&N);
            printf("%d\n",powN(N));
        }
        return 0;
    }
        
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  • 原文地址:https://www.cnblogs.com/heqinghui/p/2750482.html
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