4966-Normalized Form

Http://livearchive.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & page = show_problem & problem = 2967

First, we need to find the deepest depth, and then use parity to identify whether to perform a crossover or complement operation. An odd number indicates an operation, and an even number indicates a supplementary operation.

Note that the global variable I is used to solve the problem of repeated calculation. It is worth noting.

#include<iostream>#include<string.h>#include<stdio.h>#include<algorithm>using namespace std;char s[40000];char s2[3]="()";int i;int solve(int k,int id){    int ans=id;    for(i=++k;;i++)    {        if(s[i]==')')return ans;        else        if(s[i]=='T')        {        if(id)ans&=1;        else        ans|=1;        }        else        if(s[i]=='F')        {        if(id)ans&=0;        else        ans|=0;        }        else        if(s[i]=='(')        {        if(id)ans&=solve(i,id^1);        else        ans|=solve(i,id^1);        }    }}int main(){    int test=1;    while(cin>>s)    {        int ans=0;        int temp=0;        if(!strcmp(s,s2))break;        int len=strlen(s);        for(int j=0;j<len;j++)        {            if(s[j]=='(')temp++;            else            if(s[j]==')')                ans=max(ans,temp),temp--;        }        printf("%d. ",test++);        i=0;        temp=solve(0,ans%2);        if(temp==0)            cout<<"false"<<endl;        else            cout<<"true"<<endl;    }    //system("pause");    return 0;}