[Codeforces Round #604 (Div. 2)]D. Beautiful Sequence(贪心)

An integer sequence is called beautiful if the difference between any two consecutive numbers is equal to 1. More formally, a sequence s1,s2,…,sn is beautiful if |si−si+1|=1 for all 1≤i≤n−1.
Trans has a numbers 0, b numbers 1, c numbers 2 and d numbers 3. He wants to construct a beautiful sequence using all of these a+b+c+d numbers.
However, it turns out to be a non-trivial task, and Trans was not able to do it. Could you please help Trans?
Input
The only input line contains four non-negative integers a, b, c and d (0<a+b+c+d≤105).
Output
If it is impossible to construct a beautiful sequence satisfying the above constraints, print “NO” (without quotes) in one line.
Otherwise, print “YES” (without quotes) in the first line. Then in the second line print a+b+c+d integers, separated by spaces — a beautiful sequence. There should be a numbers equal to 0, b numbers equal to 1, c numbers equal to 2 and d numbers equal to 3.
If there are multiple answers, you can print any of them.
枚举从0，1，2，3开始
贪心，先搞完0，然后1，然后2
例如0开始，到1，如果还有0回0，否则到2.。。。。。。

#include <iostream>
#include <cstdio>
#include <cstring> 
#include <queue>
using namespace std; 

int q[100001],tot,n;

bool Try(int x,int a,int b,int c,int d) {
    tot = 0;
	if (x == 0) {
		if (!a) return tot == n; 
		a--;q[++tot] = 0;
		if(!b) return tot==n;
		b--;q[++tot] = 1;
		while(a) {
			a--;q[++tot] = 0;
			if (!b) return tot==n;
			b--;q[++tot] = 1;
		}
		if(!c) return tot==n;
		c--;q[++tot] = 2;
		while(b) {
			b--;q[++tot] = 1;
			if(!c) return tot==n;
			c--;q[++tot] = 2;
		} 
		if (!d) return tot==n;
		d--;q[++tot] = 3;
		while(c) {
			c--;q[++tot] = 2;
			if(!d) return tot==n;
			d--;q[++tot] = 3;
		}
		return tot==n;
	} 
	if (x == 1) {
		if (!b) return tot==n;
		b--;q[++tot]=1;
		while(a) {
			a--;q[++tot] = 0;
			if (!b) return tot==n;
			b--;q[++tot] = 1;
		}
		if(!c) return tot==n;
		c--;q[++tot] = 2;
		while(b) {
			b--;q[++tot] = 1;
			if(!c) return tot==n;
			c--;q[++tot] = 2;
		} 
		if (!d) return tot==n;
		d--;q[++tot] = 3;
		while(c) {
			c--;q[++tot] = 2;
			if(!d) return tot==n;
			d--;q[++tot] = 3;
		}
		return tot==n;
	}
	return 0;
}

int main() {
//	int t;
//	scanf("%d",&t);
//	while(t--) {
    int a,b,c,d; 
	scanf("%d%d%d%d",&a,&b,&c,&d);
	n = a+b+c+d;
	if(Try(0,a,b,c,d)) {
		 puts("YES");
		 for(int i = 1; i <= tot; i++) 
		   printf("%d ",q[i]);
	} else if (Try(1,a,b,c,d)) {
		 puts("YES");
		 for(int i = 1; i <= tot; i++) 
		   printf("%d ",q[i]);
	} else if(Try(0,d,c,b,a)) {
		 puts("YES");
		 for(int i = 1; i <= tot; i++) 
		   printf("%d ",3-q[i]);
	} else if (Try(1,d,c,b,a)) {
		 puts("YES");
		 for(int i = 1; i <= tot; i++) 
		   printf("%d ",3-q[i]);
	} else puts("NO");
//	}
//	}
	return 0;
}