Problem Description
Given a positive integer N, you should output the leftmost digit of N^N.
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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.
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Sample Input
2 3 4 |
Sample Output
2 2 <div style="" font-size:="" 14px;="" border-top:="" #b7cbff="" 1px="" dashed;="" font-family:="" times"="">
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.
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Author
Ignatius.L
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#include<stdio.h>
#include"algorithm"
using namespace std;
int main()
{
int testCase;
unsigned long long inNum;
scanf("%d", &testCase);
while (testCase--)
{
scanf("%lld", &inNum);
long double x = inNum*log10(inNum*1.0);
x -= unsigned long long(x);
printf("%d\n", int(pow(10.0, double(x))));
}
return 0;
}
【思路】
实数可以用科学记数法表示为: a=b*10^c,其中a是任意实数,1<=b<10,C为整数。
此题中,设输入为m,那么 m^m=k*10^n,本题要求解的就是k的整数部分,也就是int(k)
两边取对数:m*log10(m)=n+log10(k)
设x=m*log10(m)=n+log10(k),
由于1<=k<10,所以0<=log10(k)<1
n是x的整数部分。log10(k)是x的小数部分。
换句话说,log10(k)=[x的小数部分]=[x-int(x)]=[m*log10(m)-int(m*log10(m))]
所以int(k)=int(10^log10(k))=int(10^[m*log10(m)-int(m*log10(m))])
【注意事项】
由于m比较大,所以要用到unsign long long,注意输入时的%lld不要出错。