https://vjudge.net/problem/POJ-3295
题意
有五种运算符和五个参数,现在给你一个不超过100字符的算式,问最后结果是否恒为1?
分析
首先明确各运算符的意义,K(&&),A(||),N(!),E(==),C特殊判断一下。计算的话肯定是从后往前的,遇到变量进栈,遇到运算符则出栈做运算。现在的问题是各个变量的数值未知,由于只有5个变量,总共只有32总可能,所以暴力跑一遍即可,每种取值都试一次。
#include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #include<cstdio> #include<algorithm> #include<map> #include<set> #define rep(i,e) for(int i=0;i<(e);i++) #define rep1(i,e) for(int i=1;i<=(e);i++) #define repx(i,x,e) for(int i=(x);i<=(e);i++) #define X first #define Y second #define PB push_back #define MP make_pair #define mset(var,val) memset(var,val,sizeof(var)) #define scd(a) scanf("%d",&a) #define scdd(a,b) scanf("%d%d",&a,&b) #define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c) #define pd(a) printf("%d ",a) #define scl(a) scanf("%lld",&a) #define scll(a,b) scanf("%lld%lld",&a,&b) #define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c) #define IOS ios::sync_with_stdio(false);cin.tie(0) using namespace std; typedef long long ll; template <class T> void test(T a){cout<<a<<endl;} template <class T,class T2> void test(T a,T2 b){cout<<a<<" "<<b<<endl;} template <class T,class T2,class T3> void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;} template <class T> inline bool scan_d(T &ret){ char c;int sgn; if(c=getchar(),c==EOF) return 0; while(c!='-'&&(c<'0'||c>'9')) c=getchar(); sgn=(c=='-')?-1:1; ret=(c=='-')?0:(c-'0'); while(c=getchar(),c>='0'&&c<='9') ret = ret*10+(c-'0'); ret*=sgn; return 1; } const int N = 1e6+10; const int inf = 0x3f3f3f3f; const ll INF = 0x3f3f3f3f3f3f3f3fll; const ll mod = 1000000000; int T; void testcase(){ printf("Case %d:",++T); } const int MAXN = 5e5+10 ; const int MAXM = 150; const double eps = 1e-8; const double PI = acos(-1.0); int p,q,r,s,t; int sta[120]; char ss[120]; void run(){ int top=0; int len = strlen(ss); int a,b; for(int i=len-1;i>=0;i--){ if(ss[i]=='p') sta[top++]=p; else if(ss[i]=='q') sta[top++]=q; else if(ss[i]=='r') sta[top++]=r; else if(ss[i]=='s') sta[top++]=s; else if(ss[i]=='t') sta[top++]=t; else if(ss[i]=='K'){ a=sta[--top]; b=sta[--top]; sta[top++]=(a&&b); }else if(ss[i]=='A'){ a=sta[--top]; b=sta[--top]; sta[top++]=(a||b); }else if(ss[i]=='N'){ a=sta[--top]; sta[top++]=(!a); }else if(ss[i]=='C'){ a=sta[--top]; b=sta[--top]; if(a==1&&b==0) sta[top++]=0; else sta[top++]=1; }else if(ss[i]=='E'){ a=sta[--top]; b=sta[--top]; sta[top++]=(a==b); } } } bool work(){ for(p=0;p<2;p++) for(q=0;q<2;q++) for(r=0;r<2;r++) for(s=0;s<2;s++) for(t=0;t<2;t++){ run(); if(sta[0]==0) return false; } return true; } int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL // init(); while(~scanf("%s",ss)){ if(ss[0]=='0') break; if(work()) puts("tautology"); else puts("not"); } return 0; }