Poj 3295 Tautology

Mom, it turns out to be a discrete MATHEMATICAL question, a question, and an equivalent of disjunction, union, non, implication .... Thanks to the end of the final exam, it would be sad ,,,,

The general meaning of this question is to determine whether the given formula is a permanent, true, or a total of five elements. The given string is a suffix expression of the formula .. First, convert to judgment .. 2 ^ 5 in 32 cases in total, if the enumeration is false, mark the output not. Otherwise, tautology ,,,,

I will not repeat the following code...

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 110int main(){    char s[MAX], s1[MAX], str[MAX];    int t, k, i, j, flat;    int num[32][5] =    {        1,1,1,1,1,        1,1,1,1,0,        1,1,1,0,1,        1,1,1,0,0,        1,1,0,1,1,        1,1,0,1,0,        1,1,0,0,1,        1,1,0,0,0,        1,0,1,1,1,        1,0,1,1,0,        1,0,1,0,1,        1,0,1,0,0,        1,0,0,1,1,        1,0,0,1,0,        1,0,0,0,1,        1,0,0,0,0,        0,1,1,1,1,        0,1,1,1,0,        0,1,1,0,1,        0,1,1,0,0,        0,1,0,1,1,        0,1,0,1,0,        0,1,0,0,1,        0,1,0,0,0,        0,0,1,1,1,        0,0,1,1,0,        0,0,1,0,1,        0,0,1,0,0,        0,0,0,1,1,        0,0,0,1,0,        0,0,0,0,1,        0,0,0,0,0,    };    while(~scanf("%s",s))    {        flat = 0;        k = strlen(s);        if(k == 1 && s[0] == '0')            break;        for(i = 0; i < 32; i++)        {            t = 0;            for(j = 0; j < k; j++)            {                if(s[j] == 'p')                    s1[j] = num[i][0] + '0';                else if(s[j] == 'q')                    s1[j] = num[i][1] + '0';                else if(s[j] == 'r')                    s1[j] = num[i][2] + '0';                else if(s[j] == 's')                    s1[j] = num[i][3] + '0';                else if(s[j] == 't')                    s1[j] = num[i][4] + '0';                else                    s1[j] = s[j];            }            for(j = k-1; j >= 0; j--)            {                if(s1[j] == '0' || s1[j] == '1')                {                    str[t++] = s1[j];                }                else if(s1[j] == 'K')                {                    str[t-2] = ((str[t-2]-'0')&&(str[t-1]-'0'))+'0';                    t--;                }                else if(s1[j] == 'A')                {                    str[t-2] = ((str[t-2]-'0')||(str[t-1]-'0'))+'0';                    t--;                }                else if(s1[j] == 'C')                {                    str[t-2] = ((!(str[t-2]-'0'))||(str[t-1]-'0'))+'0';                    t--;                }                else if(s1[j] == 'E')                {                    str[t-2] = ((str[t-2]-'0')==(str[t-1]-'0'))+'0';                    t--;                }                else if(s1[j] == 'N')                {                    str[t-1] = (!(str[t-1]-'0'))+'0';                }            }            if(str[0] == '0')                flat = 1;        }        if(flat == 0)            printf("tautology\n");        else            printf("not\n");    }    return 0;}