hihocoder-1223-Inequalities

Enter n inequalities to find the maximum number of inequalities at the same time, data range n<=50, inequalities to the right of the data [0,1000]

Analysis: This problem has to change the idea to do, the known data not more than 1000, so enumeration 0~1000 within the number, in order to calculate how many inequalities to meet, and then update the answer, that is, a double cycle, 1000*50

However, this question does not say that must be an integer, such as x>2,x<3, if the requirement x must be an integer, then these two inequalities are not at the same time, but this question does not say, so x can be a floating-point number, then the two inequalities can be set up at the same time.

So when we do, we multiply the data one time, which is [0,2000] to do, but with the boundary problem, it becomes [ -2,2002] enumeration.

1#include <iostream>2#include <string>3#include <vector>4 using namespacestd;5 6 intN;7 strings;8 structnode{9     intFG;Ten     intx; OneNodeintfg=0,intx=0): FG (FG), X (x) {} A }; -Vector<node>v; -  the intMain () - { -Cin>>N; -      while(n--){ +         CharC; -Cin>>C; +Cin>>s; A         intx; atCin>>x; -         intFG; -         if(s=="=") fg=0; -         if(s==">") fg=1; -         if(s=="<") fg=-1; -         if(s=="<=") fg=-2; in         if(s==">=") fg=2; -Node No (FG,2*x); to V.push_back (no); +     } -     intans=0; the      for(inti=-2; i<=2002; i++){ *         intCnt=0; $          for(intj=0; J<v.size (); j + +){Panax Notoginseng             intfg=V[J].FG; -             intx=v[j].x; the             if(fg==0&AMP;&AMP;I==X) cnt++; +             Else if(fg==-1&AMP;&AMP;I&LT;X) cnt++; A             Else if(fg==1&AMP;&AMP;I&GT;X) cnt++; the             Else if(fg==-2&AMP;&AMP;I&LT;=X) cnt++; +             Else if(fg==2&AMP;&AMP;I&GT;=X) cnt++; -         } $ans=Max (ans,cnt); $     } -cout<<ans<<Endl; -}

hihocoder-1223-Inequalities