T3表达式求值

AC代码：

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>using namespace std;char str[301];char **opera=new char *[301];int value[301];int t1=0,t2=0;void p1(char s[]){opera[t1]=new char[strlen(s)+1];strcpy(opera[t1++],s);}char *pop1(){return opera[--t1];}void p2(int x){value[t2++]=x;}int pop2(){return value[--t2];}int add(int a,int b){return a+b;}int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}char s[5];inline void copy(char s1[],int pos){int i,j;for(i=pos,j=0;j<3&&s1[i];i++,j++)s[j]=s1[i];s[j]='/0';}inline int trans(char s[]){int sum=0;int i,l=strlen(s);l--;for(i=0;s[i];i++)sum+=(s[i]-'0')*(int)pow(10,l--);return sum;}int main(){int n,i,j;char s1[6];scanf("%d",&n);while(n--){t1=t2=0;scanf("%s*c",str);j=0;for(i=0;str[i];i++){if(str[i]=='(');else if(str[i]>='a'&&str[i]<='z'){copy(str,i);i+=3;p1(s);}else if(str[i]>='0'&&str[i]<='9'){int t=0;for(;str[i]&&str[i]>='0'&&str[i]<='9';i++)s1[t++]=str[i];s1[t]='/0';p2(trans(s1));i--;}else if(str[i]==','){;}else if(str[i]==')'){int a=pop2();int b=pop2();char ss[5];strcpy(ss,pop1());if(strcmp(ss,"max")==0){int x=max(a,b);p2(x);}else if(strcmp(ss,"min")==0){int x=min(a,b);p2(x);}else if(strcmp(ss,"add")==0){int x=add(a,b);p2(x);}}}printf("%d/n",value[0]);}return 0;}