AIM Tech Round 3 (Div. 1)-B. Recover the String

原题链接

B. Recover the String

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

For each string s consisting of characters '0'
and '1' one can define four integers a00, a01, a10 and a11,
where axy is the number
of subsequences of length 2 of the string s equal
to the sequence {x, y}.

In these problem you are given four integers a00, a01, a10, a11 and
have to find any non-empty string s that matches them, or determine that there is no such string. One can prove that if at least one answer exists,
there exists an answer of length no more than 1 000 000.

Input

The only line of the input contains four non-negative integers a00, a01, a10 and a11.
Each of them doesn't exceed 109.

Output

If there exists a non-empty string that matches four integers from the input, print it in the only line of the output. Otherwise, print "Impossible".
The length of your answer must not exceed 1 000 000.

Examples

input

1 2 3 4

output

Impossible

input

1 2 2 1

output

0110

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <cmath>
#define maxn 1000005
#define MOD 1000000007
#define INF 1e9
using namespace std;
typedef long long ll;

char str[maxn];
int main(){

//	freopen("in.txt", "r", stdin);
ll a, b, c, d;
scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);
a *= 2;
d *= 2;
ll k1 = sqrt(a);
ll k2 = sqrt(d);
if(k1 * (k1 + 1) != a || k2 * (k2 + 1) != d){
puts("Impossible");
return 0;
}
if(a == 0 && b == 0 && c == 0 && d == 0){
puts("0");
return 0;
}
if(a == 0 && b == 0 && c == 0 && d){
for(int i = 1; i <= k2 + 1; i++)
str[i] = '1';
puts(str+1);
return 0;
}
if(a && b == 0 && c == 0 && d == 0){
for(int i  = 1; i <= k1 + 1; i++)
str[i] = '0';
puts(str+1);
return 0;
}
if((k1+1) * (k2+1) != b + c){
puts("Impossible");
return 0;
}
k1++, k2++;
ll cc = b / k2, e = b % k2;
for(int i = 1; i <= k1 + k2; i++)
str[i] = '0';
for(int i = cc + 1; i <= cc + k2; i++)
str[i] = '1';
if(e){
str[cc+k2+1] = '1';
str[cc+k2+1-e] = '0';
}
puts(str+1);
return 0;
}