B - Brackets in Implications

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u

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 ', 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.

Sample Input

Input
4
0 1 1 0
Output
YES
(((0)->1)->(1->0))
Input
2
1 1
Output
NO
Input
1
0
Output
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\n",str2[]="NO\n";
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\n";
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;
}

XJTU Summer Holiday Test 1(Brackets in Implications-构造)的更多相关文章

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

    E. Brackets in Implications Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/conte ...

  2. codeforces #550E Brackets in Implications 结构体

    标题效果:定义集合中{0,1}\{0,1\}上的运算符"→\rightarrow",定义例如以下: 0→0=10\rightarrow 0=1 0→1=10\rightarrow ...

  3. CodeForces 550E Brackets in Implications 推理

    给出一个四个规则 0->0=1  0->1=1 1->0=0  1->1=0 我自己当时一味的去找规律,没有把式子好好推一推. 当然每个人都能想到a[n]=0是必须的 当a[n ...

  4. 「日常训练」Brackets in Implications(Codeforces Round 306 Div.2 E)

    题意与分析 稍微复杂一些的思维题.反正这场全是思维题,就一道暴力水题(B).题解直接去看官方的,很详尽. 代码 #include <bits/stdc++.h> #define MP ma ...

  5. XJTU Summer Holiday Test 1(Divisibility by Eight-8的倍数)

    C - Divisibility by Eight Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & ...

  6. CodeForces 550E Brackets in Implications(构造)

    [题目链接]:click here~~ [题目大意]给定一个逻辑运算符号a->b:当前仅当a为1b为0值为0,其余为1,构造括号.改变运算优先级使得最后结果为0 [解题思路]: todo~~ / ...

  7. Codeforces Round #306 (Div. 2)

    A. Two Substrings You are given string s. Your task is to determine if the given string s contains t ...

  8. Codeforces Round #306 (Div. 2) ABCDE(构造)

    A. Two Substrings 题意:给一个字符串,求是否含有不重叠的子串"AB"和"BA",长度1e5. 题解:看起来很简单,但是一直错,各种考虑不周全, ...

  9. Codeforces Round #306 (Div. 2) D.E. 解题报告

    D题:Regular Bridge 乱搞. 构造 这题乱搞一下即可了.构造一个有桥并且每一个点的度数都为k的无向图. 方法非常多.也不好叙述.. 代码例如以下: #include <cstdio ...

随机推荐

  1. BZOJ[Sdoi2014]数表 莫比乌斯反演

    [Sdoi2014]数表 Time Limit: 10 Sec  Memory Limit: 512 MBSubmit: 2383  Solved: 1229[Submit][Status][Disc ...

  2. CodeForces Round #403 (Div.2) A-F

    精神不佳,选择了在场外同步划水 没想到实际做起来手感还好,早知道就报名了…… 该打 未完待续233 A. Andryusha and Socks 模拟,模拟大法好.注意每次是先判断完能不能收进柜子,再 ...

  3. java IO的字节流和字符流及其区别

    1. 字节流和字符流的概念    1.1 字节流继承于InputStream    OutputStream,    1.2 字符流继承于InputStreamReader    OutputStre ...

  4. 学习webservice

    客户端测试页: WebService代码: using System; using System.Collections.Generic; using System.Linq; using Syste ...

  5. MATLAB7 + sqlitejdbc-v056.jar 访问数据库

    以下代码出错: conn=database('data.db','','','org.sqlite.JDBC','jdbc:sqlite:C:/MATLAB7/work/del_man_voice_f ...

  6. cannot load shared object file undefined symbol

    cannot load shared object file undefined symbol 场景: 共享库里引用了主程序一个符号,结构编译的时候没问题,运行时用 dlopen 打开共享库报上述错误 ...

  7. mogilefsd同步速度调优

    #查看主从mogadm settings list #一点点调试mogadm settings listmogadm settings set internal_queue_limit 500moga ...

  8. Dreamweaver安装须知

    1.断网安装,否则让你登录邮箱什么的,安装完成后退出,先不要运行: 2.将破解的文件直接复制到安装的文件的地方覆盖:(将32文件夹下的amtlib.dll复制到安装完毕的dw_cs6下,覆盖原来的am ...

  9. WebBrowser(超文本浏览框)控件默认使用IE9,IE10的方法

    C#和易语言都可以使用该方法来变更默认的的IE版本 该文是通过修改注册表的方法实现,测试的时候发现易语言本身也是采用的这种方法 操作方法 打开注册表 HKEY_LOCAL_MACHINE (or HK ...

  10. springBoot【01】

    /* 使用spring官网的 http://start.spring.io/ 来建立项目包 生成入口文件,入口文件中对类注释@SpringBootApplication,这个注释是唯一的,标明这个类是 ...