HDU 1060 Leftmost Digit (数学/大数)
2024-08-30 03:28:13
Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14954 Accepted Submission(s): 5775
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).
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
3
4
Sample Output
2
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
Recommend
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(,p);
cout<<(long long)q<<endl;
} return ;
}
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。
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