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

B. Recover the String

题目连接：

http://www.codeforces.com/contest/708/problem/B

Description

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.

Sample Input

1 2 3 4

Sample Output

Impossible

Hint

题意

你要构造一个只含有01的串，其中子序列：00有a个，01有b个，10有c个，11都d个。

如果不能，输出impossible

题解：

通过00和11，我们知道这个串里面0和1分别有多少个，然后我们就可以构造了。

假设0全部放在前面，1全部放在后面，那么如果把一个1扔到最前面，那么C就会减去当前0的个数。

然后我们就通过这个玩意儿去构造就好了。

代码

#include<bits/stdc++.h>
using namespace std;

long long f(long long x)
{
    return x*(x-1)/2;
}
long long a,b,c,d,A,B;
int flag=0;
int main()
{
    cin>>a>>b>>c>>d;
    for(long long i=0;f(i)<=a;i++)
    {
        if(f(i)==a)
        {
            for(long long j=0;f(j)<=d;j++)
            {
                if(f(j)==d)
                {
                    if(i*j==b+c)
                    {
                        A=i;
                        B=j;
                        flag=1;
                    }
                }
            }
        }
    }
    if(flag==0)
        return puts("Impossible"),0;
    if(A==0&&B==0)
        return puts("Impossible"),0;
    int sum = A+B;
    for(int i=0;i<sum;i++)
    {
        if(c>=A)
        {
            B--;
            c-=A;
            printf("1");
        }
        else
        {
            A--;
            printf("0");
        }
    }
    printf("\n");
    return 0;
}