poj 3295

写完代码发现WA，无奈之下看了讨论区中的测试数据，觉得自己弱爆了！没理解到底什么是WFF，太急于写代码了！

先把错误代码贴出来，提醒我以后一定把题目审清楚再做题！

#include <iostream>#include <stack>#include <stdlib.h>#include <stdio.h>#include <string>using namespace std;int p,q,r,s,t;int trans(char tmp){    if(tmp == '0')        return 0;    else if(tmp == '1')        return 1;    else if(tmp == 'p')        return p;    else if(tmp == 'q')        return q;    else if(tmp == 'r')        return r;    else if(tmp == 's')        return s;    else if(tmp == 't')        return t;}char transR(int tmp){    if(tmp == 0)        return '0';    else if(tmp == 1)        return '1';}int main(){    freopen("in.txt","r",stdin); //输入重定向，输入数据将从in.txt文件中读取    freopen("out.txt","w",stdout); //输出重定向，输出数据将保存在out.txt文件中    string line;    while(cin >> line)    {        if(line.length() == 1 && line[0] == '0')            break;        int flag = 1;        string origin = line;        if(line.length() == 1)            flag = 0;        for(int i = 0;i < 32;++i)        {            t = i & 1;            s = (i >> 1) & 1;            r = (i >> 2) & 1;            q = (i >> 3) & 1;            p = (i >> 4) & 1;            line = origin;            int nowSite = line.length() - 1;            while(nowSite > 0)            {                if(line[nowSite - 1] == 'N')                {                    line[nowSite - 1] = transR(!trans(line[nowSite]));                    nowSite--;                }                else                {                    int tmp1,tmp2;                    tmp1 = trans(line[nowSite - 1]);                    tmp2 = trans(line[nowSite]);                    if(line[nowSite - 2] == 'K')                    {                        line[nowSite - 2] = transR(tmp1 & tmp2);                    }                    else if(line[nowSite - 2] == 'A')                    {                        line[nowSite - 2] = transR(tmp1 | tmp2);                    }                    else if(line[nowSite - 2] == 'C')                    {                        line[nowSite - 2] = transR(!tmp1 | tmp2);                    }                    else if(line[nowSite - 2] == 'E')                    {                        if(tmp1 == tmp2)                            line[nowSite - 2] = '1';                        else                            line[nowSite - 2] = '0';                    }                    nowSite -=2;                }            }            if(line[0] == '0')            {                flag = 0;                //printf("%d\n",p);                break;            }        }        if(flag)            printf("tautology\n");        else            printf("not\n");    }    return 0;}

之后是两份网上转来的优秀的值得学习的代码，一份是用递归做的，另一份是用表达式求值做的。这个题可以算式表达式求值的变种。

#include <iostream>using namespace std;char str[101];int pos;bool judge( char str[], int value ){ pos++; switch ( str[pos] ) { case 'p': return value&1; case 'q': return (value>>1)&1; case 'r': return (value>>2)&1; case 's': return (value>>3)&1; case 't': return (value>>4)&1; case 'K': return judge(str,value)&judge(str,value); case 'A': return judge(str,value)|judge(str,value); case 'N': return !judge(str,value); case 'C': return (!judge(str,value))|judge(str,value); case 'E': return judge(str,value)==judge(str,value); default:; }}int main(){ bool mark; while ( cin >> str && str[0] != '0' ) {  mark = true;  for ( int i = 0; i < 32; i++ )  {   pos = -1;   if ( !judge(str,i) ) {    mark = false; break;   }  }  if ( mark ) cout << "tautology" << endl;  else cout << "not" << endl; } return 0;}代码2：#include <iostream>using namespace std;int Nth_Char;char str[101];bool step ( char str[101], int logicValue ){ Nth_Char++; switch ( str[Nth_Char] ) { case 'p': return logicValue & 1; case 'q': return ( logicValue >> 1 ) & 1; case 'r': return ( logicValue >> 2 ) & 1; case 's': return ( logicValue >> 3 ) & 1; case 't': return ( logicValue >> 4 ) & 1; case 'N': return !step ( str, logicValue ); case 'K': return step ( str, logicValue ) & step ( str, logicValue );   case 'A': return step ( str, logicValue ) | step ( str, logicValue ); case 'C': return !step (str, logicValue ) | step ( str, logicValue ); case 'E': return step ( str, logicValue ) == step ( str, logicValue ); default:; }}bool judge ( char str[101] ){ for ( int i = 0; i < 32; i++ ) {  Nth_Char = -1;  if ( !step ( str, i ) )   return false; } return true;}int main(){ while ( cin >> str && str[0] != '0' ) {  if ( judge ( str ) )   cout << "tautology" << endl;  else   cout << "not" << endl; } return 0;}

给一个布尔表达式，判断是不是恒真。我是枚举然后用栈做的，判断栈顶俩元素，是布尔变量就直接算，不是就接着往栈里扔。p.s.这让我想起了本学期那节逻辑课……因为样例就是p | ( !p )是恒真表达式………… /*ID: SummerDAwayPROG: 3295LANG: C++OJ: POJ.ORG*/#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std ;char a[200] ;map<char,int> f ;int top ;char Stack[200] ;char Trans( bool x ) {    if ( x )    return '1' ;    else return '0' ;}void calc( char a , char x , char y ) {    top -= 3 ;    if ( a == 'K' )    Stack[top++] = Trans( f[x] & f[y] ) ;    if ( a == 'A' )    Stack[top++] = Trans( f[x] | f[y] ) ;    if ( a == 'C' )    Stack[top++] = Trans( (!f[x]) | f[y] ) ;    if ( a == 'E' )    Stack[top++] = Trans( !( f[x] ^ f[y] ) ) ;    return ;}    bool judge( ) {    int len = strlen( a ) ;    for ( int i = 0 ; i < len ; i++ ) {        Stack[top++] = a[i] ;        bool flag = false ;        while ( top > 1 && !flag ) {            flag = true ;            char x = Stack[top-1] ;            char y = Stack[top-2] ;            if ( ( f[x] >= 0 && f[y] >= 0 ) ) {                calc( Stack[top-3] , y , x ) ;                flag = false ;            }            if ( f[x] >= 0 && y == 'N' ) {                top -= 2 ;                Stack[top++] = Trans( !f[x] ) ;                flag = false ;            }        }    }    if ( Stack[0] == '1' ) return true ;    else return false ;}                 int main() {    freopen( "3295.in" , "r" , stdin ) ;    freopen( "3295.out" , "w" , stdout ) ;    gets( a ) ;    while ( a[0] != '0' ) {        f['1'] = 1 , f['0'] = 0 ;        f['A'] = -1 , f['K'] = -1 , f['C'] = -1 , f['E'] = -1 , f['N'] = -1 ;        bool flag = true ;        for ( int t = 0 ; t < 2 ; t++ )        for ( int b = 0 ; b < 2 ; b++ )        for ( int c = 0 ; c < 2 ; c++ )        for ( int d = 0 ; d < 2 ; d++ )        for ( int e = 0 ; e < 2 ; e++ ) {            f['p'] = t , f['q'] = b , f['r'] = c , f['s'] = d , f['t'] = e ;            top = 0 ;            if ( !judge() )    {                flag = false ;            }        }        if ( flag ) printf( "tautology\n" ) ;        else printf( "not\n" ) ;        gets( a ) ;    }    return 0 ;}