hdu 1060 the leftmost digit 计算N^N最左面的一位数

最新推荐文章于 2020-03-08 17:26:37 发布

原创最新推荐文章于 2020-03-08 17:26:37 发布 · 843 阅读

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本文介绍了一个算法问题：给定正整数N，如何找出N^N的最左侧数字。通过使用对数和取幂运算，文章提供了一种高效的解决方案，并附带示例输入输出来说明算法的工作原理。

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4901 Accepted Submission(s): 1793

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.
 

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    __int64 ci;scanf("%I64d",&ci);
    while(ci--)
    {
        __int64 n;scanf("%I64d",&n);
        double l=n*log10(n+0.0);
        __int64 m=(__int64)pow(10.0,l-(__int64)l);
        printf("%I64d/n",m);
    }
    return 0;
}