【codeforces 550E】Brackets in Implications

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 ”, and the arguments and the result of the implication are written as ‘0’ (false) and ‘1’ (true). According to the definition of the implication:

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 determine if it is possible to place there brackets so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.

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 (), which means the values of arguments in the expression in the order they occur.

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.

Examples
input
4
0 1 1 0
output
YES
(((0)->1)->(1->0))
input
2
1 1
output
NO
input
1
0
output
YES
0

【题目链接】:http://codeforces.com/contest/550/problem/E

【题解】

如果是这样的
xxxxxxxxxxxxxxxxxxxx10
可以发现不管怎么样最后的答案都是0；
因为倒数第二个数字的1保证了前n-1个数字按顺序一定是1(0或1和1都是1)然后再和0混合一下就是0了；
假设倒数第二个数字不是1而是0
xxxxxxxxxxxxxxxxxxxxxxxx00
那么就在前面再找一个0
比如
xxxxxxxxxxxxxxxx0xxxxxx00
把那个0周围的按照下面处理一下
xxxxxxxxxxxxxxx(0(xxxxxx0))0
然后因为第1位是0的时候第二位是什么结果都是1
则
那个括号嵌套里的东西最终都会变成1
即
xxxxxxxxxxxxxxx10
这样就变成肯定有解的情况了;
特判一下无解的那些情况就好

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 1e5+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,a[MAXN];

void wujie()
{
    puts("NO");
    exit(0);
}

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i]);
    if (a[n]!=0)
        wujie();
    if (n==1)
    {
        puts("YES");
        printf("%d
",a[n]);
        return 0;
    }
    if (a[n-1]==1)
    {
        puts("YES");
        rep1(i,1,n)
            {
                printf("%d",a[i]);
                if (i==n)
                    puts("");
                else
                    printf("->");
            }
        return 0;
    }
    if (n==2)
        wujie();
    int pos = n-2;
    while (pos>=1 && a[pos]==1) pos--;
    if (pos==0) // 111100
        wujie();
    else
    {
        puts("YES");
        rep1(i,1,pos-1)
            printf("%d->",a[i]);
        printf("(0->(");
        rep1(i,pos+1,n-2)
            printf("%d->",a[i]);
        printf("0))->0");
        puts("");
    }
    return 0;
}