Codeforces Round #306 (Div. 2) E. Brackets in Implications

分类讨论就行了。。。。想要搞成0的话最后一位一定要是0，然后前面有00,01,10,11四种情况，每种情况都想一下就行了。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 100005
#define eps 1e-12
#define mod 1000003
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#pragma comment(linker, "/STACK:102400000,102400000")
#define pii pair<int, int> 
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

int a[maxn];
char s[maxn];

void work()
{
	int n;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
	if(n == 1) {
		if(a[1] == 0) printf("YES\n0\n");
		else printf("NO\n");
		return;
	}

	if(a[n] == 1) {
		printf("NO\n");
		return;
	}

	if(a[n-1] == 1) {
		printf("YES\n");
		for(int i = 1; i <= n; i++) printf("%d%s", a[i], i == n ? "\n" : "->");
	}
	else {
		if(n == 2) printf("NO\n");
		else if(a[n-2] == 0) {
			printf("YES\n");
			for(int i = 1; i <= n-3; i++) printf("%d%s", a[i], i == n ? "\n" : "->");
			printf("(0->0)->0\n");
		}
		else {
			int ok = 0, pos;
			for(int i = n-2; i >= 1; i--) if(a[i] == 0) {
				ok = 1;
				pos = i;
				break;
			}
			if(!ok) printf("NO\n");
			else {
				printf("YES\n");
				for(int i = 1; i < pos; i++) printf("%d->", a[i]);
				printf("(0->(");
				for(int i = pos+1; i <= n-2; i++) printf("%d->", a[i]);
				printf("0))->0\n");
			}
		}
	}

}

int main()
{
	work();

	return 0;
}