XJTU Summer Holiday Test 1(Brackets in Implications-构造)
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 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\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
- 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 ...
- 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 ...
- 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 ...
- 「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 ...
- 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 & ...
- 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~~ / ...
- 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 ...
- 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, ...
- 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
- java 复习整理(三 修饰符)
访问控制修饰符 Java中,可以使用访问控制符来保护对类.变量.方法和构造方法的访问.Java支持4种不同的访问权限. 默认的,也称为default,在同一包内可见,不使用任何修饰符. 私有的,以pr ...
- python类中两个列表实例如何相加或相减
如下 import numpy a = [1, 2, 3, 4] b = [5, 6, 7, 8] a_array = numpy.array(a) b_array = numpy.array(b) ...
- python常用20库
python核心库和统计 简述 1. Requests.最着名的http库由kenneth reitz编写.这是每个python开发人员必备的. 2. Scrapy.如果您参与webscraping, ...
- 使用 padding-bottom 设置高度基于宽度的自适应
我们在做移动端列表,通常会做到图文列表,列表是自适应的.当列表中有图片,图片的宽度是随着列表宽的变化而变化,我们为了在图片宽度变化的时候做到图片的不变形,所有采用以下办法. 本文章只讲语法 html ...
- vue学习之环境配置
最近在学习vue,就顺手记录一下... 1. 安装 nodejs https://nodejs.org -->注:安装LTS的(LTS为长期稳定版本) 在cmd中输入 node -v 如果显 ...
- WCF 小程序案例以及序列化的使用
using System;using System.Collections.Generic;using System.Linq;using System.Runtime.Serialization;u ...
- hdu 5101(思路题)
Select Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Subm ...
- Dell Inspiron 7520 安装Ubuntu 14.04 LTS
我的电脑是Dell Inspiron 7520,之前用的Windows 7, 装了虚拟机,再在虚拟机里面装Ubuntu, 电脑一直卡顿,一怒之下,升级了内存,直接16G,然后,还是卡顿,CPU是i5的 ...
- Code+ B 汀博尔【二分答案】
时间限制:C/C++ 1秒,其他语言2秒空间限制:C/C++ 262144K,其他语言524288K64bit IO Format: %lld 题目描述 有 n 棵树,初始时每棵树的高度为 Hi,第 ...
- Python的扩展接口[2] -> 动态链接库DLL[0] -> 动态链接库及辅助工具
动态链接库 / Dynamic Link Library 目录 动态链接库简介 函数封装DLL 组件对象模型COM 如何判断.dll文件是COM还是DLL 辅助工具 1 动态链接库简介 / DLL I ...