ACM刷题之Codeforces————Make a Square

最新推荐文章于 2020-10-26 07:42:17 发布

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最新推荐文章于 2020-10-26 07:42:17 发布

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给定一个正整数 nn，不包含前导零。每次操作可以删除一个数字，使得结果仍为正整数且无前导零。找出使 nn 成为某个正整数平方所需的最少操作数，或报告不可能。如果不可能将 nn 转换为正整数的平方，则输出 -1。例如，8314 经过删除 3 和 4 后变成 8181，即 99 的平方。对于 333333，无法变为正整数的平方，所以答案是 -1。

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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 $n$ ( $1 \leq n \leq 2 \cdot 10^{9}$ ). 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
inputCopy
8314
outputCopy
2
inputCopy
625
outputCopy
0
inputCopy
333
outputCopy
-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

用暴力深搜就好了.

下面是ac代码：

#include<bits/stdc++.h>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;
int le,i, j ,k, e;
char c[100];

void dfs(int x, int val, int step) {
	int b = c[x] - '0';
	if (x == le - 1) {
		//走到底
		int c = sqrt(val);
		if (c * c == val && val != 0) {
			e = min(step + 1, e);
		} 
		val = val*10 + b;
		c = sqrt(val);
		if (c * c == val && val != 0) {
			e = min(step, e);
		} 
		
		return ;
	}
	if (b != 0 || val != 0) {
		dfs(x + 1, val * 10 + b, step);
	}
	dfs(x + 1, val, step + 1);
	
}

int main()
{
	//freopen("f:/input.txt", "r", stdin);
	cin>>c;
	le = strlen(c);
	e = INF;
	dfs(0, 0, 0);
	if (e == INF) {
		cout<<-1<<endl;
		return 0;
	}
	cout<<e<<endl;
}