AIM Tech Round 3 (Div. 2) D. Recover the String (构造)

本文介绍了一个基于四个整数重构特定二进制字符串的问题,并提供了一种解决方案。通过数学方法确定了字符串中0和1的分布规律,从而实现了字符串的有效重建。

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D. 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 a00a01a10 and a11, where axy is the number ofsubsequences of length 2 of the string s equal to the sequence {x, y}.

In these problem you are given four integers a00a01a10a11 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 than1 000 000.

Input

The only line of the input contains four non-negative integers a00a01a10 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




/* ***********************************************
┆  ┏┓   ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃       ┃ ┆
┆┃   ━   ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃       ┃ ┆
┆┃   ┻   ┃ ┆
┆┗━┓ 马 ┏━┛ ┆
┆  ┃ 勒 ┃  ┆      
┆  ┃ 戈 ┗━━━┓ ┆
┆  ┃ 壁     ┣┓┆
┆  ┃ 的草泥马  ┏┛┆
┆  ┗┓┓┏━┳┓┏┛ ┆
┆   ┃┫┫ ┃┫┫ ┆
┆   ┗┻┛ ┗┻┛ ┆
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <bitset>
using namespace std;

#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")


int f(int x)
{
    if(x&1)return (x-1)/2*x;
    else return x/2*(x-1);
}


int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	ios::sync_with_stdio(0);
	cin.tie(0);
	int a,b,c,d;
	scanf("%d%d%d%d",&a,&b,&c,&d);
	int flag = 0;
	int A=0,B=0;
	for(int i=0;f(i)<=a;i++)
    {
        if(f(i)==a)
        {
            for(int j=0;f(j)<=d;j++)
            {
                if(f(j)==d)
                {
                    if(i*j==b+c)
                    {
                        A = i;
                        B = j;
                        flag = 1;
                    }
                }
            }
        }
    }
    if(!flag)
    {
        printf("Impossible\n");
    }
    else
    {
        int sum = A + B;
        for(int i=1;i<=sum;i++)
        {
            if(c>=A)
            {
                c -= A;
                B--;
                printf("1");
            }
            else
            {
                A--;
                printf("0");
            }
        }
        printf("\n");
    }


	return 0;
}




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