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。

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <sstream>
10 #include <queue>
11 #include <typeinfo>
12 #include <fstream>
13 #include <map>
14 #include <stack>
15 using namespace std;
16 #define INF 100000
17 typedef long long ll;
18 const int maxn=1010;
19
20 int main()
21 {
22     int t;
23     scanf("%d",&t);
24     double sum;
25     while(t--){
26         ll n;
27         scanf("%I64d",&n);
28         sum=n*log10(n);
29         sum-=(ll)sum;
30         printf("%I64d\n",(ll)pow(10.0,sum));
31     }
32     return 0;
33 }