大数运算

B题 HDU1060

Leftmost Digit

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).

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.
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解题思路：
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1、m=n^n,要取m的最高位；
2、log10(m)=log10(n^n);
3、m=10^(n*log10(n));
4、设n*log10(n)=a+b; (a为整数部分，b为小数部分)
5、则m=10^(a+b)=(10^a)*(10^b);
6、由于10^a的最高位是1，不影响结果，所以m的最高位只需考虑10^b；
7、最后对10^b的结果取整；（由于0<b<10,所以1<10^b<10，即为所求结果）
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题目代码：
 
 

  
  
#include <bits/stdc++.h>
typedef long long LL;
using namespace std;
int main(){
    int n,t;
    double x,y;
    LL result;
    cin>>t;
    while(t--){
        cin>>n;
        x=n*log10(n);
        y=LL(n*log10(n));
        x=x-y;
        result=(LL)pow(10.0,x);
        cout<<result<<endl; 
    }
    return 0;
}