hdu 1060 Leftmost Digit-优快云博客

本文介绍了一种方法来计算给定大数的幂次方后的左起数字，通过取对数和利用数学特性简化计算过程。

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m=n^n;两边同取对数，得到，log10(m)=n*log10(n);再得到，m=10^(n*log10(n));
然后，对于10的整数次幂，第一位是1，所以，第一位数取决于n*log10(n)的小数部分

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13967 Accepted Submission(s): 5339

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

代码：

#include <stdio.h>
#include <math.h>
int main()
{
    double test;
    double a;
    double c;
    __int64 b;
    __int64 d;
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf",&test);
    
        a=test*log10(test);
        b=(__int64)a;
        c=a-b;
        d=(__int64)(pow(10,c));
        printf("%I64d\n",d);
    }
    return 0;
}