HDU-1060(简单数学)

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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).

 

Output

For each test case, you should output the leftmost digit of N^N.

 

Sample Input

2

3

4

 

Sample Output

2

2

 

 

分析：求n^n的最前一位数。有log10(n^n)=n*log10(n)=c,令a为c的整数部分，b为c的小数部分，则n^n=10^c=10^a*10^b,即10^b=10^(c-a)，10^b的整数部分只有一位就是n^n的最左位。

 

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <ctime>
 #include <iostream>
 #include <algorithm>
 #include <set>
 #include <vector>
 #include <sstream>
 #include <queue>
 #include <typeinfo>
 #include <fstream>
 #include <map>
 #include <stack>
 using namespace std;
 #define INF 100000
 typedef long long ll;
 const int maxn=;
 
 int main()
 {
     int t;
     scanf("%d",&t);
     double sum;
     while(t--){
         ll n;
         scanf("%I64d",&n);
         sum=n*log10(n);
         sum-=(ll)sum;
         printf("%I64d\n",(ll)pow(10.0,sum));
     }
     return ;
 }