AIM Tech Round 3 (Div. 2)D. Recover the String（贪心+字符串）

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 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 exceed109.

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

比赛时，一直纠结怎么判断Impossible和构造字符串，胡乱交了一发，wa了，好SB。

当然先是利用a00和a11求1和0的字符串个数g1和g0，如果a00或a11等于1，这时候还要根据a10和a01判断a00或者a11是等于1还是0.

在求完a00和a11的个数后，判断合不合理，C(g0,2)=a00, C(g1,2)=a11, C(g1+g0,2) = a00+a01+a10+a11即成立。

然后就是构造字符串了。

设初始的字符串个数为0000001111111......

然后就是贪心的构造字符串了。考虑该输出某一位时，如果在这一位时，a10>=g0,那么我可以把1挪到最前面，输出1，这时候后面剩下的字符串需要构造的a10-=g0,剩余的1的个数g1--。否则直接输出0，后面剩余字符串需要构造a01-=g1,剩余的0的个数g0--.

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 typedef long long ll;
 int main()
 {
     ll a00,a01,a10,a11;
     scanf("%I64d%I64d%I64d%I64d",&a00,&a01,&a10,&a11);
     ll g0,g1;
     g0 = (+sqrt(+*a00))/;
     g1 = (+sqrt(+*a11))/;
     if(a00+a01+a10+a11==)
     {
          printf("0\n");
          return ;
     }
     if(g0==||g1==)
     {
         if(g0==)
         {
             if(a01||a10) g0 = ;
             else g0 = ;
         }
         if(g1==)
         {
             if(a01||a10) g1 = ;
             else g1 = ;
         }
     }
     if(g0*(g0-)!=*a00||g1*(g1-)!=*a11)
     {
         printf("Impossible\n");
         return ;
     }
     ll gz = g0+g1;
     if(gz*(gz-)/!=(a00+a01+a10+a11))
     {
         printf("Impossible\n");
         return ;
     }
     while(g0+g1)
     {
         if(a10>=g0)
         {
             printf("");
             a10 -= g0;
             g1--;
         }
         else
         {
             printf("");
             a01 -= g1;
             g0--;
         }
     }
     printf("\n");
     return ;
 }