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.
The only line of the input contains four non-negative integers a00, a01, a10 and a11. Each of them doesn't exceed 109.
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.
1 2 3 4
Impossible
1 2 2 1
0110
首先从a00和a11可以求出0和1的数量；
假设目前为0000...001111..；
显然a01=num0*num1，a10=0;
那么我们移动一个1去左边，a10++,a01--;但总数还是不变；
那么合理的解必须是a10+a01=num0*num1;
值得注意的是：当00或11=0时，0或1可能有1个也可能没有；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int a00, a01, a10, a11; int main()
{
//ios::sync_with_stdio(0);
rdint(a00); rdint(a01); rdint(a10); rdint(a11);
int k = 1, r = 1;
while (k*(k - 1) / 2 < a00)k++;
while (r*(r - 1) / 2 < a11)r++;
// cout << k << ' ' << r << endl;
int fg = 1;
if (k*(k - 1) / 2 != a00 || r * (r - 1) / 2 != a11) {
cout << "Impossible" << endl; return 0;
} else if (a00 == 0 && a11 == 0) {
if (a01 == 0 && a10 == 0)cout << 0 << endl;
else if (a01 == 0 && a10 == 1)cout << "10" << endl;
else if (a01 == 1 && a10 == 0)cout << "01" << endl;
else cout << "Impossible" << endl;
return 0;
}
else if (a00 == 0) {
if (a01 == 0 && a10 == 0) {
while (r--)cout << '1'; }
else if (a01 + a10 == r) {
while (a10--)cout << '1';
cout << 0;
while (a01--)cout << '1';
}
else fg = 0;
}
else if (a11 == 0) {
if (a10 == 0 && a01 == 0) {
while (k--)cout << '0';
}
else if (a10 + a01 == k) {
while (a01--)cout << 0;
cout << 1;
while (a10--)cout << 0;
}
else fg = 0;
} else {
if (a01 + a10 == k * r) {
while (a01) {
while (r > a01) {
cout << 1;
r--;
}
cout << 0;
k--; a01 -= r;
}
while (r--)cout << 1;
while (k--)cout << 0;
}
else fg = 0;
}
if (fg == 0)cout << "Impossible" << endl;
return 0;
}