1 /*  2   3   4 L (>=60), the lower bound of the qualified grades --   5 that is, only the ones whose grades of talent and virtue are both not below  6 this line will be ranked;   7   8 and H (<100), the higher line of qualification  9  10  11 ,those with both grades not below this line are considered as the "sages", 12 and will be ranked in non-increasing order according to their total grades.  13  14 Those with talent grades below H but virtue grades not are cosidered as  15 the "noblemen", and are also ranked in non-increasing order according to  16 their total grades, but they are listed after the "sages".  17  18  19 Those with both grades below H,  20 but with virtue not lower than talent are considered as the "fool men". 21 They are ranked in the same way but after the "noblemen".  22  23  24 The rest of people whose grades both pass the L line are ranked  25 after the "fool men". 26  27  28  29 */ 30  31 #include <string.h> 32 #include <map> 33 #include <iomanip> 34 #include <algorithm> 35 #include <vector> 36 #include <stdio.h> 37 using namespace std; 38  39 struct peo 40 { 41     char num[9]; 42     int all,t,v; 43     int id; 44 }; 45  46 int high; 47  48 bool cmp(peo a,peo b) 49 { 50     if(a.id==b.id) 51     { 52         if(a.all == b.all) 53         { 54             if(a.v == b.v ) 55                 return (strcmp(a.num,b.num)<0); 56             else return a.v > b.v; 57         } 58         else return a.all > b.all; 59     } 60     else return a.id<b.id; 61  62 } 63  64 int main() 65 { 66  67     int i,j,n,low,v,t; 68     char num[9]; 69     while(scanf("%d%d%d",&n,&low,&high)!=EOF) 70     { 71  72         vector<peo> VP; 73         for(i=0;i<n;i++) 74         { 75             getchar(); 76             scanf("%s %d %d",num,&v,&t); 77             if(t>=low && v>=low) 78             { 79                 peo pp; 80                 strcpy(pp.num,num); 81                 pp.v=v; 82                 pp.t=t; 83                 pp.all=v+t; 84                 if(v>=high && t>=high) 85                     pp.id=1; 86                 else if(v>= high && t<high) 87                     pp.id=2;     88                 else if(v< high && t< high && v>=t) 89                     pp.id=3; 90                 else  91                     pp.id=4; 92  93                 VP.push_back(pp); 94             } 95         } 96  97         sort(VP.begin(),VP.end(),cmp); 98  99 100         printf("%d\n",VP.size());101         102 103         for(i=0;i<VP.size();i++)104             printf("%s %d %d\n",VP[i].num,VP[i].v,VP[i].t);105         106 107 108     }109 110 111 112 113     return 0;114 }