C. Make a Square

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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=y2 for some positive integer y.

Input

The first line contains a single integer n (1≤n≤2⋅109). 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.

题意:

给你一个字符串,让你删除最少的字符串个数,使其剩余的字符串代表的数字没有前导0,并且是一个数的平方数。

思路:

因为字符串的长度是 2e9 ,我们知道 y的最大范围 sqrt(2e9) 那么我们显然可以枚举每一个y,把他的平方数转为字符串(长度最大为9),去和给定的字符串进行匹配,检测是否可以是给定字符串的子序列,并维护满足子序列的不同字符个数的最小值就是答案。

时间复杂度 O( 9 * sqrt(2e9 ) )

细节见代码:

#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const ll inf=1e18+7;
/*** TEMPLATE CODE * * STARTS HERE ***/
string S(ll n){stringstream ss;string s;ss<<n;ss>>s;return s;}
ll N(string s){stringstream ss;ll n;ss<<s;ss>>n;return n;}
string a;
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
 
cin>>a;
ll ans=inf;
int len=a.length();
ll y=1ll;
while(1)
{
ll x=y*y;
string str=S(x);
int slen=str.size();
if(slen>len)
break;
int id=0;
for(int i=0;i<len;++i)
{
if(a[i]==str[id])
{
id++;
}
}
if(id==slen)
{
ans=min(ans,1ll*len-slen);
}
y++;
}
 
if(ans==inf)
{
ans=-1;
}
cout<<ans<<endl;
 
return 0;
}
 
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
 
 
 

C Make a Square Educational Codeforces Round 42 (Rated for Div. 2) (暴力枚举,字符串匹配)的更多相关文章

  1. Educational Codeforces Round 42 (Rated for Div. 2) C

    C. Make a Square time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  2. Educational Codeforces Round 42 (Rated for Div. 2) E. Byteland, Berland and Disputed Cities

    http://codeforces.com/contest/962/problem/E E. Byteland, Berland and Disputed Cities time limit per ...

  3. Educational Codeforces Round 42 (Rated for Div. 2) D. Merge Equals

    http://codeforces.com/contest/962/problem/D D. Merge Equals time limit per test 2 seconds memory lim ...

  4. Educational Codeforces Round 42 (Rated for Div. 2)F - Simple Cycles Edges

    http://codeforces.com/contest/962/problem/F 求没有被两个及以上的简单环包含的边 解法:双联通求割顶,在bcc中看这是不是一个简单环,是的话把整个bcc的环加 ...

  5. Educational Codeforces Round 42 (Rated for Div. 2)

    A. Equator(模拟) 找权值的中位数,直接模拟.. 代码写的好丑qwq.. #include<cstdio> #include<cstring> #include< ...

  6. Educational Codeforces Round 42 (Rated for Div. 2) B

    B. Students in Railway Carriage time limit per test 2 seconds memory limit per test 256 megabytes in ...

  7. Educational Codeforces Round 42 (Rated for Div. 2) A

    A. Equator time limit per test 2 seconds memory limit per test 256 megabytes input standard input ou ...

  8. D. Merge Equals(from Educational Codeforces Round 42 (Rated for Div. 2))

    模拟题,运用强大的stl. #include <iostream> #include <map> #include <algorithm> #include < ...

  9. D Merge Equals Educational Codeforces Round 42 (Rated for Div. 2) (STL )

    D. Merge Equals time limit per test2 seconds memory limit per test256 megabytes inputstandard input ...

随机推荐

  1. 造题inginging

    造个题 模拟+sort+贪心 蚕丛及鱼凫,造题何茫然 U74939 小歪被抓走了 代码(不知道对不对哦) #include<bits/stdc++.h> using namespace s ...

  2. Retrofitting Analysis

    Retrofitting Analysis To figure out the process of retrofitting[1] objective updating, we do the fol ...

  3. golang context学习记录2

    上篇文章中,我们已经学习了使用context实现控制多个goroutine的退出. 本文将继续介绍如何使用context实现超时情况下,让多个goroutine退出. 例子 首先,启动3个gorout ...

  4. 汽车Lin总线特点

    串行通信:线间干扰小,节省线束,传输距离长 单线传输:增强的ISO 9141 (ISO 15765-1),总线电压基于VBAT 最高速率:20kbit/s 单主多从结构,无需仲裁:主节点同时包含主任务 ...

  5. 第二章 SpringCloud之Eureka-Server服务发现组件

    1.Eureka简介 文档:https://cloud.spring.io/spring-cloud-netflix/spring-cloud-netflix.html ############### ...

  6. C#规范整理·多线程\异步\并行\任务

    有一个领域的工作处理起来几乎总是最棘手的,这就是多线程编码.多线程编码是所有开发人员前进途中的一个坎,现在,该是尝试克服它的时候了. 1.区分异步和多线程应用场景 先看一个例子 private voi ...

  7. PyCharm安装+破解

    PyCharm 是一款功能强大的 Python 编辑器,具有跨平台性,鉴于目前最新版 PyCharm 使用教程较少,为了节约时间,来介绍一下 PyCharm 在 Windows下是如何安装的. 这是 ...

  8. 【HANA系列】SAP ECLIPSE中创建ABAP项目失败原因解析

    公众号:SAP Technical 本文作者:matinal 原文出处:http://www.cnblogs.com/SAPmatinal/ 原文链接:[HANA系列]SAP ECLIPSE中创建AB ...

  9. Unity中的动画系统和Timeline(5) Timeline

    在前面的动画,都是控制单独的物体,比如说控制一个角色的运动.而Timeline,可以对多个物体实施动画,形成过场动画,或者电影效果.比如,很多赛车游戏比赛开始前都会播放一段开场动画,围绕自己车的几个方 ...

  10. session 的理解