Codeforces 550E Brackets in Implications

本文探讨了如何通过合理地添加括号和->操作符到一个由0和1组成的序列中,使得最终的逻辑表达式的值为0。特别关注了在不同情况下如何构造这些表达式,并提供了一个具体的算法实现。

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E. Brackets in Implications
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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 v1v2 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
Copy
4
0 1 1 0
output
Copy
YES
(((0)->1)->(1->0))
input
Copy
2
1 1
output
Copy
NO
input
Copy
1
0
output
Copy
YES
0

题意:给了你0 1的序列和四种->的规则,问能否通过加入->和()使得最后结果为0

分析:由于只有1->0才为0,所以最后一个数一定是0。前面的数如果是1,结果也一定为0,因为无论0还是1指向1都是1。前面的数如果是0,在这个0之前找最近的一个0,因为0->1=1,所以这之间的1运算完还是1,可以把它们看成一个整体,不断加括号就可以实现。而这个整体1之前无论是0还是1运算完都是1,所以一定能构造出来。但是如果之前没有0了,就不能构造。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int a[maxn],s;
int n;
bool fun()
{
    for(int i = n - 2; i >= 1; i--)
    {
        if(a[i] == 0)
        {
            s = i;
            return true;
        }
    }
    return false;
}
int main()
{
    cin>>n;
    for(int i = 1; i <= n; i++)
        cin>>a[i];
    if(n == 1 && a[n] == 0)
    {
        printf("YES\n0\n");
    }
    else if(a[n] == 1)
        puts("NO");
    else
    {
        if(a[n - 1] == 1)
        {
            puts("YES");
            int first = 0;
            for(int i = 1; i <= n; i++)
            {
                if(first)printf("->%d",a[i]);
                else
                {
                    first = 1;
                    printf("%d",a[i]);
                }
            }
            printf("\n");
        }
        else
        {
            if(fun())
            {
                puts("YES");
                int first = 0;
                for(int i = 1; i <= n; i++)
                {
                    if(first)printf("->");
                    else first = 1;
                    if(s == i)
                    {
                        printf("(%d",a[i]);
                       for(int j = i + 1; j <= n - 1; j++)
                            printf("->(%d",a[j]);
                       for(; i <= n - 1; i++)
                        printf(")");
                        i--;
                    }
                    else printf("%d",a[i]);
                }
            }
            else
                puts("NO");
        }
    }
    return 0;
}


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