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
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Sample Output
2
2
c=log10(n^n)=nlog10(n),a是c的整数部分,b是c的小数部分,a+b=c;n^n=10^c=10^a*10^b,所以10^b=10^c/10^a=10^(c-a), 10^b的整数部分是答案。 |
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define INF 0x3f3f3f3f #define Max int(5e5+10) int aa[Max]; int main() { int t,n,i,j,c; scanf("%d",&t); while(t--) { scanf("%d",&n); ll a=n*log10(n); double b=n*log10(n)-a; printf("%d\n",(int)pow(10.0,b)); } return 0; }