判断一个数是否能整开方,perfect square

int m,n;
m=sqrt(n);
m*m==n?yes:no;

https://www.codewars.com/kata/56269eb78ad2e4ced1000013/solutions/java

 

public class NumberFun {
  public static long findNextSquare(long sq) {
      long root = (long) Math.sqrt(sq);
      return root * root == sq ? (root + 1) * (root + 1) : -1;
  }
}

 

来源：https://www.cnblogs.com/ukzq/p/11427108.html