acm steps(Leftmost Digit)

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2483    Accepted Submission(s): 1088 

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

234

 

Sample Output

22
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.
没思路直接翻题解了，如下
m=n^n;两边同取对数，得到，log10(m)=n*log10(n);再得到，m=10^(n*log10(n));
然后，对于10的整数次幂，第一位是1，所以，第一位数取决于n*log10(n)的小数部分
总之，log很强大啊，在求一个数的位数上，在将大整数化成范围内的整数上
我的AC代码
#include<iostream>#include<string>#include<cstring>#include<iomanip>#include<cstdio>#include<algorithm>#include<vector>#include<cmath>#include<set>using namespace std;typedef long long ll;typedef vector<int> vi;typedef set<int> si;int main(){   int n;   double x;   cin>>n;   while(n>0)   {       n--;      cin>>x;       double a=x*log10(x);       ll b=static_cast<ll>(a);       a-=b;       cout<<static_cast<int>(pow(10,a))<<endl;   }   return 0;}
/////没注意a转换成int会类型溢出，WA了几次，改成longlong就好了