Problem

Recall that a subsequence of a string is any string obtained by removing some subset of characters from the string, for instance “string”, “sing”, “i” and “sg” are all subsequences of “string”. If the same subsequence can be obtained in exactly t different ways, by removing different subsets of characters, we say that the subsequence occurs t times.

Jingfei wants to create a nonempty bit string that has the following properties:

1. the subsequence 00 occurs a  times,

• the subsequence 01 occurs b times,

• the subsequence 10 occurs c  times, and

• the subsequence 11 occurs d    times.

However, Jingfei does not know how to create such a string – or whether it is even possible. Could you help her?

Input

The input consists of a single line with four integers a, b, c, and d (0≤a,b,c,d≤1e9).

Output

Output a bit string that satisfies the given requirements. If there are several solutions, output any one of them. If there are no solutions, output “impossible”.

3 4 2 1

01001

5 0 0 5

impossible

題解 ：不需要考慮00、11，根據a個00和d個11來算出來需要的0的個數x，1的個數y。

           找答案字串時，湊01，這樣10也滿足條件。

           設q = b / y，w = b % y。這樣先輸出時，輸出q個0，輸出y-w個1，這樣就保證了有q*（y-w）個01，如果w == 0，表示剛好能夠用上所有的1來組成01，再把剩餘的輸出1和0就可以了。但是如果q != x，即不需要把0全部輸出，那樣q*（y-w）> b，所以把0剩餘提前到1前面，eg：現在有3個0，3個1，我們只需要7個01，按著這種想法，如果輸出001110（程式碼如果沒考慮）或者是000111，這樣就會多答案或者少答案，所以應該是001101，所以要判斷一下餘下的，把0調前一個，讓正好能夠是7個01就可以了，相應的最後的輸出0個數要-1。

        其次加上特判，a、b、c、d分別為0和部分為0的時候的特判。（自己好菜。。%隊友）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll a,b,c,d,abc;
    scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
    if(a==0&&b==0&&c==0&&d==0)
    {
        printf("0\n");
        return 0;
    }
    ll x,y,i;
    x = (1 + sqrt(1+8*a)) / 2;
    if(x*(x-1)/2 != a)
    {
        printf("impossible\n");
        return 0;
    }
    y = (1 + sqrt(1 + 8*d)) / 2;
    if(y*(y-1)/2 != d)
    {
        printf("impossible\n");
        return 0;
    }
    if(a == 0&&d == 0)
    {
        if(b == 1&&c == 0)
        {
            printf("0");
            printf("1\n");
        }
        else if(c == 1&&b==0)
        {
            printf("1");
            printf("0\n");
        }
        else printf("impossible\n");
        return 0;
    }
    if(b==0&&c==0)
    {
        if(a==0)
        {
            for(i=1;i<=y;i++) printf("1");
            printf("\n");
        }
        else if(d==0)
        {
            for(i=1;i<=x;i++) printf("0");
            printf("\n");
        }
        else printf("impossible\n");
        return 0;
    }
    if(x*y != b+c)
    {
        printf("impossible\n");
        return 0;
    }
    ll j,q,w;
    q = b / y;
    w = b % y;
    for(i=1;i<=q;i++) printf("0");
    for(i=1;i<=y-w;i++) printf("1");
    if(q!=x)
    printf("0");
    for(i=1;i<=w;i++) printf("1");
    for(i=1;i<=x-q-1;i++) printf("0");
    return 0;
}