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).

 

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】：

【代码】：

#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
",ans);
    }
    return 0;
}