Codeforces1132A——Regular Bracket Sequence（水题）

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探讨如何判断一组特定括号序列，包括((、()、)(和))，能否通过合理排列形成规范括号序列。文章分析了括号配对条件，并提供了算法实现。

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Regular Bracket Sequence

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

A string is called bracket sequence if it does not contain any characters other than “(” and “)”. A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters “+” and “1” into this sequence. For example, “”, “(())” and “()()” are regular bracket sequences; “))” and “)((” are bracket sequences (but not regular ones), and “(a)” and “(1)+(1)” are not bracket sequences at all.

You have a number of strings; each string is a bracket sequence of length $2$ . So, overall you have $c n t 1$ strings “((”, $c n t 2$ strings “()”, $c n t 3$ strings “)(” and $c n t 4$ strings “))”. You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length $2 (c n t 1 + c n t 2 + c n t 3 + c n t 4)$ . You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either.

Input

The input consists of four lines, $i$ -th of them contains one integer $cnt_i$ $0≤cnt_i≤10^9)$ .

Output

Print one integer: $1$ if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, $0$ otherwise.

Examples

input

output

input

output

input

output

Note

In the first example it is possible to construct a string “(())()(()((()()()())))”, which is a regular bracket sequence.

In the second example it is possible to construct a string “”, which is a regular bracket sequence.

Solve

你有四种括号，"(("，"()"，")("，"))"，给出四种括号的数量，问能否将这些括号以某种顺序连接起来，使得每个左括号都有与之匹配的右括号。

观察可以发现：如果要完全匹配，()是不会对结果产生影响的，然后可以推出一个关系式： $cnt1=2×cnt3+cnt4,cnt4=2×cnt3+cnt1cnt1=2\times cnt3+cnt4,cnt4=2\times cnt3+cnt1$

即：当 $cnt3≠0,cnt1=cnt4≠0cnt3\neq0,cnt1=cnt4\neq0$ 或 $c n t 3 = 0, c n t 1 = c n t 2$ 时，才会完全匹配

Code

/*************************************************************************

	 > Author: WZY
	 > School: HPU
	 > Created Time:   2019-03-15 10:34:53
	 
************************************************************************/
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstring>
#include <limits.h>
#include <iostream>
#include <algorithm>
#include <random>
#include <iomanip>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <random>
#define ll long long
#define ull unsigned long long
#define lson o<<1
#define rson o<<1|1
#define ms(a,b) memset(a,b,sizeof(a))
#define SE(N) setprecision(N)
#define PSE(N) fixed<<setprecision(N)
#define bug cerr<<"-------------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
#define LEN(A) strlen(A)
const double E=exp(1);
const double eps=1e-9;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int maxn=1e6+10;
const int maxm=1e3+10;
const int moha=19260817;
const int inf=1<<30;
const ll INF=1LL<<60;
using namespace std;
inline void Debug(){cerr<<'\n';}
inline void MIN(int &x,int y) {if(y<x) x=y;}
inline void MAX(int &x,int y) {if(y>x) x=y;}
inline void MIN(ll &x,ll y) {if(y<x) x=y;}
inline void MAX(ll &x,ll y) {if(y>x) x=y;}
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cerr<<arg<<"";Debug(rest...);}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);cin.tie(0);
	cout.precision(20);
	#ifndef ONLINE_JUDGE
	    freopen("in.txt", "r", stdin);
	    freopen("out.txt", "w", stdout);
		srand((unsigned int)time(NULL));
	#endif
	int cnt1,cnt2,cnt3,cnt4;
	while(cin>>cnt1>>cnt2>>cnt3>>cnt4)
	{
		if(cnt1+cnt4==0)
		{
			if(cnt3)
				cout<<0<<endl;
			else
				cout<<1<<endl;
			continue;
		}
		if(cnt1==cnt4)
			cout<<1<<endl;
		else
			cout<<0<<endl;
	}
	#ifndef ONLINE_JUDGE
	    cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s.\n";
	#endif
	return 0;
}