给定一个整数n，我们需要找出从数字中删除多少个数字，使其成为一个完美的平方。
例子 ：
Input : 8314
Output: 81 2
Explanation: If we remove 3 and 4 number becomes 81 which is a perfect square.
Input : 57
Output : -1
想法是生成所有可能的子序列，并使用设置的位返回最佳字符串。假设我们有一个字符串8314。并且使用置位，我们形成了所有可能的子序列，即
8、3、83、1、81、31、831、4、84、34、834、14、814、314、8314。
形成所有可能的子序列后，我们检查哪一个是理想的正方形。然后我们返回一个最小长度的理想平方数。
在上面的示例中，三个完美的平方分别是1 4和81，因此答案将是81，因为81的最大长度为2。
C++
// C++ program to find required minimum digits
// need to remove to make a number perfect squre
#include
using namespace std;
// function to check minimum number of digits
// should be removed to make this number
// a perfect square
int perfectSquare(string s)
{
// size of the string
int n = s.size();
// our final answer
int ans = -1;
// to store string which is perfect square.
string num;
// We make all possible subsequences
for (int i = 1; i < (1 << n); i++) {
string str = "";
for (int j = 0; j < n; j++) {
// to check jth bit is set or not.
if ((i >> j) & 1) {
str += s[j];
}
}
// we do not consider a number with leading zeros
if (str[0] != '0') {
// convert our temporary string into integer
int temp = 0;
for (int j = 0; j < str.size(); j++)
temp = temp * 10 + (int)(str[j] - '0');
int k = sqrt(temp);
// checking temp is perfect square or not.
if (k * k == temp) {
// taking maximum sized string
if (ans < (int)str.size()) {
ans = (int)str.size();
num = str;
}
}
}
}
if (ans == -1)
return ans;
else {
// print PerfectSquare
cout << num << " ";
return n - ans;
}
}
// Driver code
int main()
{
cout << perfectSquare("8314") << endl;
cout << perfectSquare("753") << endl;
return 0;
} Java
// Java program to find required minimum digits
// need to remove to make a number perfect squre
import java.io.*;
import java.lang.*;
public class GFG {
// function to check minimum
// number of digits should
// be removed to make this
// number a perfect square
static int perfectSquare(String s)
{
// size of the string
int n = s.length();
// our final answer
int ans = -1;
// to store string which
// is perfect square.
String num = "";
// We make all possible subsequences
for (int i = 1; i < (1 << n); i++) {
String str = "";
for (int j = 0; j < n; j++) {
// to check jth bit is set or not.
if (((i >> j) & 1) == 1) {
str += s.charAt(j);
}
}
// we do not consider a number
// with leading zeros
if (str.charAt(0) != '0') {
// convert our temporary
// string into integer
int temp = 0;
for (int j = 0; j <
str.length(); j++)
temp = temp * 10 +
(int)(str.charAt(j) - '0');
int k = (int)Math.sqrt(temp);
// checking temp is perfect
// square or not.
if (k * k == temp) {
// taking maximum sized string
if (ans < (int)str.length()) {
ans = (int)str.length();
num = str;
}
}
}
}
if (ans == -1)
return ans;
else {
// print PerfectSquare
System.out.print(num + " ");
return n - ans;
}
}
// Driver code
public static void main(String args[])
{
System.out.println(perfectSquare("8314"));
System.out.println(perfectSquare("753"));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)Python3
# C++ program to find required minimum
# digits need to remove to make a
# number perfect squre
import math
# function to check minimum number of
# digits should be removed to make
# this number a perfect square
def perfectSquare(s) :
# size of the string
n = len(s)
# our final answer
ans = -1
# to store string which is
# perfect square.
num = ""
# We make all possible subsequences
for i in range(1, (1 << n)) :
str = ""
for j in range(0, n) :
# to check jth bit is
# set or not.
if ((i >> j) & 1) :
str = str + s[j]
# we do not consider a number
# with leading zeros
if (str[0] != '0') :
# convert our temporary
# string into integer
temp = 0;
for j in range(0, len(str)) :
temp = (temp * 10 +
(ord(str[j]) - ord('0')))
k = int(math.sqrt(temp))
# checking temp is perfect
# square or not.
if (k * k == temp) :
# taking maximum sized
# string
if (ans < len(str)) :
ans = len(str)
num = str
if (ans == -1) :
return ans
else :
# print PerfectSquare
print ("{} ".format(num), end="")
return n - ans
# Driver code
print (perfectSquare("8314"))
print (perfectSquare("753"));
# This code is contributed by
# manishshaw1.C#
// C# program to find required minimum digits
// need to remove to make a number perfect squre
using System;
class GFG {
// function to check minimum
// number of digits should
// be removed to make this
// number a perfect square
static int perfectSquare(string s)
{
// size of the string
int n = s.Length;
// our final answer
int ans = -1;
// to store string which
// is perfect square.
string num = "";
// We make all possible subsequences
for (int i = 1; i < (1 << n); i++) {
string str = "";
for (int j = 0; j < n; j++) {
// to check jth bit is set or not.
if (((i >> j) & 1) == 1) {
str += s[j];
}
}
// we do not consider a number
// with leading zeros
if (str[0] != '0') {
// convert our temporary
// string into integer
int temp = 0;
for (int j = 0; j < str.Length; j++)
temp = temp * 10 + (int)(str[j] - '0');
int k = (int)Math.Sqrt(temp);
// checking temp is perfect
// square or not.
if (k * k == temp) {
// taking maximum sized string
if (ans < (int)str.Length) {
ans = (int)str.Length;
num = str;
}
}
}
}
if (ans == -1)
return ans;
else {
// print PerfectSquare
Console.Write(num + " ");
return n - ans;
}
}
// Driver code
public static void Main()
{
Console.WriteLine(perfectSquare("8314"));
Console.WriteLine(perfectSquare("753"));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)PHP
> $j) & 1)
{
$str = $str.$s[$j];
}
}
// we do not consider a
// number with leading zeros
if ($str[0] != '0')
{
// convert our temporary
// string into integer
$temp = 0;
for ($j = 0; $j < strlen($str); $j++)
$temp = $temp * 10 +
(ord($str[$j]) - ord('0'));
$k = (int)(sqrt($temp));
// checking temp is perfect
// square or not.
if (($k * $k) == $temp)
{
// taking maximum sized string
if ($ans < strlen($str))
{
$ans = strlen($str);
$num = $str;
}
}
}
}
if ($ans == -1)
return $ans;
else
{
// print PerfectSquare
echo ($num." ");
return ($n - $ans);
}
}
// Driver code
echo (perfectSquare("8314"). "\n");
echo (perfectSquare("753"). "\n");
// This code is contributed by
// Manish Shaw (manishshaw1)
?>输出 ：
81 2
-1