HDU1060 Leftmost Digit 【数学】

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12897    Accepted Submission(s): 4934

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.
 

 

因为N^N = A * 10^X (科学计数法,A即为所求最高位)，两边同时取对数得N * log10(N) = log10(A) + X, 得 log10(A) = N * log10(N) - X, 得 A = 10^(N*log10(N) - X), 其中X为N^N结果的位数-1，即[log10(N*N)], 中括号表示取整，所以最终结果为A = 10^(N*log10(N) - [N*log10(N)]).需要注意的是N^N的位数要用__int64类型存储。

#include <stdio.h>#include <math.h>int main(){    int t, n;    double ans;    scanf("%d", &t);    while(t--){        scanf("%d", &n);        ans = n * log10(n);        ans -= (__int64)ans;        ans = pow(10, ans);        printf("%d\n", (int)ans);    }    return 0;}