Description
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character 
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
When there are brackets, we first calculate the expression in brackets. For example,
For the given logical expression 
Input
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbers a1, a2, ..., an (
Output
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide with a1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
- Expressions "0", "1" (without the quotes) are correct.
- If v1, v2 are correct, then v1->v2 is a correct expression.
- If v is a correct expression, then (v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed 106.
If there are multiple possible answers, you are allowed to print any of them.
Sample Input
4 0 1 1 0
YES (((0)->1)->(1->0))
2 1 1
NO
1 0
YES 0
这题要用构造法，
n=1特判
考虑n>=2的情况
不难发现最后那个数必须是0。否则无解（由于1和不论什么数左运算结果为1）
倒数第二个数若为1，则（..1）->0 =0
否则倒数第二个数为0：
此时若倒数第三个数为0 (...(0->0))->0=0 (0->0)
否则倒数第三个数为1 (...(1->0))->0 因为(..1)->0 =0 所以把1右运算 ..->(1->0)=..->0 想让结果为1，则..=0
考虑前面有1个0
(...->(0->(1->1->..->1->0))->0 = (..->(0->0))->0=(..->1)->0=1->0 =0
有解
否则前面均为1
1->1->1->1->0->0 不管怎么括都无解
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<string>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int a[MAXN],n;
char str1[]="YES
",str2[]="NO
";
string logic(string s1,string s2)
{
string p=""+s1+"->"+s2+"";
p="("+p+")";
return p;
}
string itos(int x)
{
if (x) return string("1");
return string("0");
}
string logic(int i,int j)
{
string p(itos(a[i]));
Fork(k,i+1,j)
{
p+="->"+itos(a[k]);
}
p="("+p+")";
return p;
}
int main()
{
// freopen("B2.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;
For(i,n) scanf("%d",&a[i]);
if (a[n]==1)
{
cout<<str2;
return 0;
}
if (n==1)
{
cout<<str1<<"0
";
return 0;
}
if (a[n-1]==1)
{
string p;
p=logic(logic(1,n-1),"0");
cout<<str1<<p<<endl;
return 0;
}
if (a[n-1]==0)
{
ForD(i,n-2)
if (a[i]==0)
{
string p=logic(i+1,n-1);
p=logic("0",p);
if (i>1) p=logic(logic(1,i-1),p);
p=logic(p,"0");
cout<<str1<<p<<endl;
return 0;
}
}
cout<<str2;
return 0;
}