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;
}