CodeForces 708B Recover the String

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 ofsubsequences 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 than1 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

可以根据a00和a11分别确定0和1的个数

注意边界情况：a00是0，0的个数可以为1，要根据a01的值判断

在判断impossible方面，首先a00和a11必须是n*(n+1)/2的形式，

其次a01+a10=n0*n1

然后构造一个符合条件的字符串

先把所有1放前面，所有0放后面，根据a01的个数将0向前移动

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>

using namespace std;
int a00,a01,a10,a11;
int n0,n1;
int main()
{
   scanf("%d%d%d%d",&a00,&a01,&a10,&a11);
    int b1=0,b2=0;
    n0=1;n1=1;
    if(a00==0&&a01==0&&a10==0&&a11==0)
    {
        printf("0
");
        return 0;
    }
    while(b1<a00) {b1+=n0;n0++;}
    while(b2<a11) {b2+=n1;n1++;}

    if(a00==0&&a01==0&&a10==0) n0=0;
    if(a11==0&&a01==0&&a10==0) n1=0;

    if(b1!=a00||b2!=a11||(a01+a10!=n1*n0))
    {
        printf("Impossible
");
        return 0;
    }
    if(a00==0&&a10==0&&a01==0)
    {
        for(int i=0;i<n1;i++)
        printf("1");
        printf("
");
        return 0;
    }
    if(a11==0&&a10==0&&a01==0)
    {
        for(int i=0;i<n0;i++)
        printf("0");
        printf("
");
        return 0;
    }

    int num1=a01/n1;
    int num2=a01%n1;
    for(int i=0;i<num1;i++)
        printf("0");
    for(int i=0;i<n1-num2;i++)
        printf("1");
      if(num1!=n0)
        printf("0");
    for(int i=0;i<num2;i++)
        printf("1");
    for(int i=0;i<n0-num1-1;i++)
        printf("0");
    printf("
");
    return 0;
}