HDU 1060 Leftmost Digit （数学/大数）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14954 Accepted Submission(s): 5775


[align=left]Problem Description[/align]
Given a positive integer N, you should output the leftmost digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the leftmost digit of N^N.

[align=left]Sample Input[/align]

2
3
4

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
Ignatius.L

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没有思路，直接百度的

m=n^n;
linag两边同时log10
log10(m)=log10(n^n)=n*log10(n);
所以，m=10^(n*log10(n));
m的最左边一位由n*log10(n)的小数部分决定。m=10^((n*log10(n))的整数部分)+10^((n*log10(n))的小数数部分);10的整数次方肯定是1000……，最左边一位由10^((n*log10(n))的整数部分)决定。

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
int T;cin>>T;
while(T--)
{
double n;
cin>>n;
double m=n*log10(n);
double p=m-(long long)m;
double q=pow(10,p);
cout<<(long long)q<<endl;
}

return 0;
}

以后遇到大数，可以考虑log，考虑转换成指数形式，
另外大数数学的题，不要用int，用 long long，第一次就是用int，所以wa了一次。