noip的等价表达式

来源：互联网发布：ghost恢复linux 编辑：程序博客网时间：2024/04/30 22:22

#include<cstdio>#include<cmath>#include<cstdlib>#include<algorithm>#include<cstring>#include<string>#include<iostream>using namespace std;const int q[5]={1,2,3,4,5};const int inf=0x3f3f3f3f;string s,ch;int n,ans[5];int number(string &s){int ans=0,len=s.length();for (int i=0;i<len;i++) ans=ans*10+s[i]-'0';return ans;}int pow(int a,int b){int ans=1;for (int i=1;i<=b;i++) ans=ans*a;return ans;}int opt(string &s){int len=s.length(),num=0;bool b[4];b[0]=b[1]=b[2]=false;b[3]=true;for (int i=0;i<len;i++) {if (s[i]=='(') num++;if (s[i]==')') num--;if (num) continue;if (s[i]=='+'||s[i]=='-') b[0]=true;if (s[i]=='*') b[1]=true;if (s[i]=='^') b[2]=true;}for (int i=0;i<=3;i++) if (b[i]) return i;}int pan(string s,int a){int len=s.length();if (len==0) return 0;int tt=1,ans,last=0,num=0,op=opt(s);if (op==0){ans=0;for (int i=0;i<len;i++){if (s[i]=='(') num++;if (s[i]==')') num--;if (num) continue;if (s[i]!='+'&&s[i]!='-') continue;ans=ans+tt*pan(s.substr(last,i-last),a);tt= s[i]=='+' ? 1:-1;last=i+1;}ans=ans+tt*pan(s.substr(last,len-last),a);return ans;}if (op==1){ans=1;for (int i=0;i<len;i++){if (s[i]=='(') num++;if (s[i]==')') num--;if (num) continue;if (s[i]!='*') continue;ans=ans*pan(s.substr(last,i-last),a);last=i+1;} ans=ans*pan(s.substr(last,len-last),a);return ans;}if (op==2) {ans=-1;for (int i=0;i<len;i++){if (s[i]=='(') num++;if (s[i]==')') num--;if (num) continue;if (s[i]!='^') continue;if (ans==-1) ans=pan(s.substr(last,i-last),a);else ans=pow(ans,pan(s.substr(last,i-last),a));last=i+1;}ans=pow(ans,pan(s.substr(last,len-last),a));return ans;}if (s[0]=='(') return pan(s.substr(1,len-2),a);if (s[0]=='a') return a;return number(s);}void read(string &s){string y;getline(cin,y);int len=y.length();s.clear();for (int i=0;i<len;i++) if (y[i]!=' ') s.push_back(y[i]);}void debug(){int tmp;read(s);cin>>tmp;getchar();int ans=pan(s,tmp);printf("%d\n",ans);}int main(){//debug();read(s);for (int i=0;i<5;i++) ans[i]=pan(s,q[i]);cin>>n;getchar();bool o;for (int i=0;i<n;i++){read(ch);o=true;for (int j=0;j<5;j++) if (pan(ch,q[j])!=ans[j]) o=false;if (o) printf("%c",65+i);}return 0;}/*(a+1)^220a*a+2*a+1a^2+2*a+2-1-1+1*/

总结

1：对于这种字符串题，递归处理是一种好的方法

2：对于这些对时间复杂度要求不是特别高但是，考验代码实现技巧的这一类字符串题，可以通过string类型来简化我们的程序

3：string一系列函数待填坑

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