HDU1060 Leftmost Digit(大数相乘与log10的巧妙运用) （C++题解）

B题 HDU1060

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22158    Accepted Submission(s): 8577

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.

解题思路：

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）

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;
}