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  • Boolean Expressions

    Boolean Expressions
    Time Limit: 1000MS   Memory Limit: 30000K
         

    Description

    The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: 
    Expression: ( V | V ) & F & ( F | V )

    where V is for True, and F is for False. The expressions may include the following operators: ! for not , & for and, | for or , the use of parenthesis for operations grouping is also allowed. 

    To perform the evaluation of an expression, it will be considered the priority of the operators, the not having the highest, and the or the lowest. The program must yield V or F , as the result for each expression in the input file. 

    Input

    The expressions are of a variable length, although will never exceed 100 symbols. Symbols may be separated by any number of spaces or no spaces at all, therefore, the total length of an expression, as a number of characters, is unknown. 

    The number of expressions in the input file is variable and will never be greater than 20. Each expression is presented in a new line, as shown below. 

    Output

    For each test expression, print "Expression " followed by its sequence number, ": ", and the resulting value of the corresponding test expression. Separate the output for consecutive test expressions with a new line. 

    Use the same format as that shown in the sample output shown below. 

    Sample Input

    ( V | V ) & F & ( F| V)
    !V | V & V & !F & (F | V ) & (!F | F | !V & V)
    (F&F|V|!V&!F&!(F|F&V))

    Sample Output

    Expression 1: F
    Expression 2: V
    Expression 3: V
    分析:表达式求值,中缀转后缀再求值;
       当前操作符是(时,直接入栈;)时,一直出栈到(结束;!时,直接进栈,待进数时,直接出栈;其余按优先级来即可;
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <bitset>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #define rep(i,m,n) for(i=m;i<=n;i++)
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define vi vector<int>
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define ll long long
    #define pi acos(-1.0)
    #define pii pair<int,int>
    #define sys system("pause")
    const int maxn=1e5+10;
    const int N=5e4+10;
    const int M=N*10*10;
    using namespace std;
    inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
    inline void umax(ll &p,ll q){if(p<q)p=q;}
    inline void umin(ll &p,ll q){if(p>q)p=q;}
    int n,m,k,t,ret[maxn],cas;
    char a[maxn],b[maxn],c[maxn];
    void gao()
    {
        int top1=0,top2=0;
        for(int i=0;a[i];i++)
        {
            if(a[i]==' ')continue;
            if(a[i]=='(')c[++top2]=a[i];
            else if(a[i]==')')
            {
                while(c[top2]!='(')b[top1++]=c[top2--];
                top2--;
            }
            else if(a[i]=='!')
            {
                c[++top2]=a[i];
            }
            else if(a[i]=='&')
            {
                while(c[top2]=='&'||c[top2]=='!')b[top1++]=c[top2--];
                c[++top2]=a[i];
            }
            else if(a[i]=='|')
            {
                while(top2&&c[top2]!='(')b[top1++]=c[top2--];
                c[++top2]=a[i];
            }
            else
            {
                b[top1++]=a[i];
                while(c[top2]=='!')b[top1++]=c[top2--];
            }
        }
        while(top2)b[top1++]=c[top2--];
        b[top1]=0;
    }
    int main()
    {
        int i,j;
        while(gets(a))
        {
            gao();
            int top=0;
            for(i=0;b[i];i++)
            {
                if(b[i]=='V')ret[++top]=1;
                else if(b[i]=='F')ret[++top]=0;
                else if(b[i]=='!')ret[top]^=1;
                else if(b[i]=='&')
                {
                    ret[top-1]&=ret[top];
                    top--;
                }
                else
                {
                    ret[top-1]|=ret[top];
                    top--;
                }
            }
            printf("Expression %d: %c
    ",++cas,"FV"[ret[1]]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dyzll/p/6407510.html
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