NEFU 66 leftmost number

Leftmost number

Time limit 1000 ms Memory limit 65536 K

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

The question is very understandable, that is, the number on the leftmost side of N ^ N, the range of N is so large, it is obvious that you don't need to think about it directly. What should we do? Of course, the formula is used:

10 ^ (N * lg (N)-[N * lg (n)]) = POW (10, N * log10 (N)-(INT) (N * lg (N )))

I want to explain how this formula comes from:

Set n ^ n = A0 * 10 ^ m + A1 * 10 ^ m-1) +...

A0, A1. .. is the coefficient of the corresponding digit. m is the number of digits, for example, 4 ^ 4 = 256, a0 = 2, a1 = 5, a2 = 6, M = 3;

Obviously, a0 is the leftmost number, that is, what we want. Then we will ask A0;

A0 * 10 ^ m <= n ^ n <(a + 1) * 10 ^ m

The logarithm of the two keys is as follows:

M + lga0 <= nlgn <m + lg (A0 + 1 ).................................... (#)

That is, lga0 <= nlgn-M <lg (A0 + 1)

Therefore, a0 <= 10 ^ (nlgn-m) <A0 + 1, a0 is an integer, A0 + 1 is also an integer, the two are adjacent, and 10 ^ (nlgn-m) if the value of A0 + 1 is not obtained, the 10 ^ (nlgn-m) is rounded down to A0, so a0 = [10 ^ (nlgn-m)];

At this step, M is unknown, and m is to be obtained. You need to analyze A0 first:

Because 1 <= A0 <= 9, 0 <= lg (A0) <1, while 0 <lg (A0 + 1) <= 1;

Let's look at the (#) formula:

M + lga0 <= nlgn <m + lg (A0 + 1), respectively get the integer: M = [nlgn], so m is obtained.

So a0 = [10 ^ (N * lgn-m)] = [10 ^ (N * lgn-[N * lg (n)];

Code:

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

Now let's expand it, that is, ask n! The formula for the number of digits:

LG (1) + lg (2) + lg (3) + ......... LG (n) + 1

After using the real sum, convert int to + 1. You can do the NEFU 65th question.

NEFU 66 leftmost number