How does javascript prove that every digit is the same decimal number is not a complete square

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I believe all of you have done "AbigB Problem". This question is an enhanced version of it.

Enter two integers A, B for A B, for example 3, 6 for 666. All you have to do is root "A B". If the root is followed by an integer, output the number after the root, otherwise output "We donot have SPJ!"

Problem-solving ideas

Obviously, this question is for us to judge whether "A B" is a complete square. From a sense of judgment, a number like 666 ⋯666 is generally not a complete square. Now let's prove it.

Prove

So the above situation is not a complete square.

Therefore, it is not a complete square number for "A 3 and 7".

Therefore, for the "A B" of any three digits or more, it must not be a complete square.

Code / / by Concyclics#include using namespace std;int main () {int Astami B; cin > > A > > B; if (Block0) {puts ("0"); return 0;} if (Basilic1) {puts ("1") {puts ("1"); return 0 } if (Block4) {puts ("2"); return 0;} if (Block9) {puts ("3"); return 0;} puts ("We donot have SPJ!"); return 0 } after reading this, the article "how to prove that everyone has the same decimal number is not a complete square number" has been introduced. If you want to master the knowledge points of this article, you still need to practice and use it. If you want to know more about the article, please follow the industry information channel.