Nbut 1118 Marisa ' s affair (sort statistics, water)

Test instructions: Each row gives a name and an int value, and the name can be repeated. Required for the same personal name statistics int value, the largest first output, if the same, then according to the number of person names appear, if the same, then the name of the dictionary sequence.

Train of thought: After entering all sorts, write a comparison function to pass in the sort.

1#include <bits/stdc++.h>2 using namespacestd;3 4 structnode5 {6     Charnam[ A];7     intlik;8     intTim;9}a[5500];Ten  Onemap<string,int> mapp2;//Name Mapping to int is better handled A  - intCMP (node Ll,node RR) - { the     if(ll.lik<Rr.lik) -         return 0; -     Else if(ll.lik>Rr.lik) -         return 1; +  -     if(ll.tim<Rr.tim) +         return 0; A     Else if(ll.tim>Rr.tim) at         return 1; -  -  -     if(strcmp (Ll.nam, Rr.nam) >0) -         return 0; -     Else    return 1; in } -  to intMain () + { -     //freopen ("Input.txt", "R", stdin); the  *     //Sort by probability, sort by time, and then order by dictionary.  $ Panax Notoginseng     intn, like; -     Chars[ A]; the     stringtemp; +map<string,int>:: iterator it; A  the      while(cin>>N) +     { -Memset (A,0,sizeof(a)); $ mapp2.clear (); $  -         intk=0; -          for(intI=0; i<n; i++) the         { -scanf"%s%d",s,&Like );Wuyitemp=s; theit=Mapp2.find (temp); -  Wu             if(It==mapp2.end ())//new here. -             { Aboutmapp2[temp]=K; $  -a[k].lik=Like ; -a[k].tim=1; - strcpy (A[k].nam, s); Ak++; +             } the             Else -             { $a[it->second].lik+=Like ; thea[it->second].tim++; the             } the         } theSort (a,a+k,cmp); -printf"%d\n", k); in          for(intI=0; i<k; i++) theprintf"%s%d%d\n", A[i].nam, A[i].lik, a[i].tim); the     } About     return 0; the}

AC Code

Nbut 1118 Marisa ' s affair (sort statistics, water)