Leftmost digit (solution Report)

Leftmost Digit

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)

Total submission (s): 364 accepted submission (s): 198

Problem descriptiongiven a positive integer N, you should output the leftmost digit of N ^ n.

Inputthe 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 ).

Outputfor each test case, You shoshould 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.

The question is to enter N and find the highest digit of N ^ n. 1 <= n <= 1,000,000,000

It is estimated that we have no idea about the N range. The number N is indeed too large. If you want to calculate the result, it will time out even if it does not overflow.

I have been struggling with this question for a long time. At the prompt of the students, the AC is displayed.

The question is converted in this way.

First, use scientific notation to represent n ^ n = A * 10 ^ X; for example, n = 3; 3 ^ 3 = 2.7*10 ^ 1;

The rightmost number we require is (INT) A, that is, the integer part of;

OK, and then obtain the base-10 logarithm lg (N ^ n) = lg (A * 10 ^ X) on both sides );

Simplify N * lg (n) = lg (A) + X;

Continue N * lg (N)-x = lg ()

A = 10 ^ (N * lg (N)-x );

Now only X is unknown. If n is used to represent X, this problem is solved.

Because X is the number of digits n ^ n. For example, N ^ n = 1200 => X = 3; X is actually an integer down from lg (N ^ N), which is expressed as [lg (N ^ n)].

OK a = 10 ^ (N * lg (N)-[lg (N ^ n)]); then (INT) A is the answer.

Code:

#include<iostream>#include<math.h>int main(){int n,m;std::cin>>n;while(n--){std::cin>>m;long double t = m*log10(m*1.0);t -= (__int64)t;__int64 ans = pow((long double)10, t);std::cout<<ans<<std::endl;}return 0;}