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.
Meaning of the questions:
to give you a string, so you delete the fewest number of strings, so the rest of the string represents a number without a leading zero, and is the square of a number.
Idea:
because the length of the string is 2e9, we know that y the maximum range sqrt (2e9) it is clear that we can enumerate every y, his square number into a string (maximum length of 9), and given to string matching may be a subsequence of detecting whether a given string, and maintain a different number of characters to meet the minimum sequence is the answer.
The time complexity of O (9 * sqrt (2e9))
See details Code:
#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';
}
}
}