HDU 1060 Leftmost Digit【log10/求N^N的最高位数字是多少】
Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19010 Accepted Submission(s): 7507
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).
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.
Author
Ignatius.L
【分析】:![](https://images2017.cnblogs.com/blog/1184092/201801/1184092-20180125154326084-1449532615.png)
![](https://images2017.cnblogs.com/blog/1184092/201801/1184092-20180125154326084-1449532615.png)
【代码】:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int MOD = 7; typedef vector<LL> vec; typedef vector<vec> mat; int main() { int n,t,ans; double tmp; cin>>t; while(t--){ cin>>n; tmp=n*log10(1.0*n); tmp=tmp-(__int64)tmp; ans=(int)(pow(10.0,tmp)); printf("%d\n",ans); } return 0; }