Title Description:
If the number of squares at the end of a number equals this number, then the number is called the self-hold number.
Obviously, 5 and 6 are a self-observing number (5x5=25 6x6=36), 25x25=625 76x76=5776, so 25 and 76 are two-bit self-guarding.
Enter a number to determine whether the number is a self-hold number.
The idea of solving problems: it is obviously undesirable to use the method of "finding the square of a number and then intercepting the last corresponding digit", because the computer cannot represent an oversized integer. So we take the idea of multiplying by large numbers to store the square of this number. Then we can judge whether the latter of the square term is equal to the original number.
Import Java.util.Scanner; PublicFinalclassDemo { Public Static void Main(string[] args) {//TODO auto-generated method stubScanner Scanner =NewScanner (System.inch); System. out. println ("Please num:");intn = scanner.nextint (); System. out. println (Isautomorphicnum (n)); } Public StaticBooleanIsautomorphicnum(intNUM) {if(Num <0){return false; }if(num = =0){return true; } String s = string.valueof (num);int[] Number =New int[S.length ()]; for(intI=0; I<s.length (); i++) {number[s.length ()-i-1] = S.charat (i)-' 0 ';//"12345", 5,4,3,2,1}int[] result = multi (number, number); String str =""; for(inti=result.length-1; i>=0; i--) {str + = Result[i]; } String str1 = num +"";if(Str1.equals (Str.substring (Str.length ()-str1.length ()))) {//Determine if the number is self-guarding return true; }return false; }///The following method implements the multiplication of large numbers, saving each bit of the results in the result array Public Static int[]Multi(intNum1[],intNum2[]) {intLen1 = Num1.length;intLen2 = Num2.length;int[] result =New int[Len1 + len2]; for(intI=0; i<len1;i++) { for(intj=0; j<len2;j++) {Result[i+j] + = num1[i]*num2[j]; } } for(intI=0; i<result.length-1; i++) {if(Result[i] >Ten) {result[i+1] + = result[i]/Ten; Result[i]%=Ten; } }returnResult }}
Judging self-guarding number