codeforces 962 C. Make a Square

Description

You are given a positive integer $n$, written without leading zeroes (for example, the number 04 is incorrect).

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer $n$ to make from it the square of some positive integer or report that it is impossible.

An integer $x$ is the square of some positive integer if and only if $x$ = $y^2$ for some positive integer $y$ .

Input

The first line contains a single integer $n$ ( 1≤n≤2⋅10$^9$ ). The number is given without leading zeroes.

Output

If it is impossible to make the square of some positive integer from n, print -1. In the other case, print the minimal number of operations required to do it.

Examples

Input
8314
Output
2

Input
625
Output
0

Input
333
Output
-1

Note

In the first example we should delete from $8314$ the digits $3$ and $4$. After that $8314$ become equals to $81$, which is the square of the integer $9$.

In the second example the given $625$ is the square of the integer $25$, so you should not delete anything.

In the third example it is impossible to make the square from $333$, so the answer is $-1$.

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参考了下面的博客

https://www.cnblogs.com/a249189046/p/8821300.html

题意：给一个整数，每次删除一个位，要求删除后的首位不能是0，问多少次操作以后变为平方数。
思路：百度了一下开方函数的复杂度，没有看的很懂但是其中有一个说法是：

• 取对数->乘0.5->取指数

其中第一步和第三步可以用硬件实现，这样下来复杂度就是O(1)的，所以用开方函数来检查一个整数是否是平方数是可以的。
由于输入最多有10位，搜索起来最坏的情况要做10!次，这是不能接受的，这是由于重复搜索导致的，所以使用set记录搜索过的数字，把次数降到了2^10次（巨大的优化）

#include<stdio.h>
#include<math.h>
#include<queue>
#include<set>
using std::queue;
using std::set;
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define mabs(x) ((x)>0?(x):(0-(x)))
#define N_max 44722
int n;
int a10[N_max];
queue<int> a;
set<int> b;
int r(int x, int i) { 
    return x / a10[i + 1]*a10[i] + x%a10[i]; 
}

int bfs(int len) {
    int next,temp;
    for (int k = 0; k < len; ++k) {
        int t = a.size(), x;
        for (int i = 0; i < t; ++i) {
            //遍历上一层产生的数
            x = a.front(); a.pop();
            for (int q = 0; q < len - k; ++q) {//搜索去掉每一位后产生的数
                if (q>0&&q ==len-k-1 && ((x%a10[q]) / a10[q - 1]) == 0)
                //判断是否在首位以及同时第二位是否是0
                    continue;
                next = r(x, q);//生成下一个数
                if (next <= 0)continue;//特判
                //利用set检查该数是否出现过
                if (b.find(next) == b.end())b.insert(next);
                else { continue; }
                //利用sqrt检查该数是否是平方数
                temp = sqrt(next);
                if (temp*temp == next)
                    return k;
                //入队列，为下一层bfs做准备
                a.push(next);
            }
        }
    }
}
int main() {
    scanf("%d", &n);

    int len = 0,temp;
    temp = sqrt(n);
    if (temp*temp == n) {
        printf("0"); return 0;
    }
    //计算位数，保存在len里
    for (temp = n; temp > 0; ++len)temp /= 10;
    //计算10, 100, 1000,...方便取模
    a10[0] = 1;
    for (int i = 1; i < 15; ++i) {
        a10[i] = a10[i - 1] * 10;
    }
    a.push(n);
    temp = 1 + bfs(len);
    if(temp<len)printf("%d",temp);
    else printf("-1");
}