Olympic ranking problems score: 2

time limit:1 seconds Memory Limit:32 MB Special Questions:Whether Submit:62 Solution:31 tags

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Title Description

According to the requirements, to the country ranking.

Input format

There are multiple sets of data.
The first line gives the number of countries N, the number of countries required to rank m, the country number from 0 to N-1.
The second line starts with n rows for the number of Olympic gold medals, medals, and population (million) for a given country or region.
The next line gives the M country number.

Output

There are 4 ways of sorting: Gold medal total Medal total medal population Proportion medal population proportion
Best ranking and final rankings for each country
Format: rank: Rank Way
If you have the same final rank, the ranking is the lowest ranking, for the ranking method, total gold medal < medal total < gold population ratio < medal population ratio
If there is a parallel ranking, that is, if the total number of gold medals is 100,90,90,80. The ranking is 1,2,2,4.
Add a blank line after each set of data.

Sample input

4 4
4 8 1
6 6 2
4 8 2
2 12 4
0 1 2 3
4 2
8 10 1
8 11 2
8 12 3
8 13 4
0 3

Sample output

1:3
1:1
2:1
1:2

1:1

1:1

1 //at nine degrees WA, but other Ojac2 3  4#include <iostream>5#include <algorithm>6 using namespacestd;7 structE8 {9     intnum,golde,model,people;Ten intoder[4]; One}buf[100000]; A   - BOOLcmp0 (E a,e b) - { the returna.num<B.num; - } -   - BOOLCmp1 (E a,e b) + { - returnA.golde>B.golde; + } A   at BOOLCMP2 (E a,e b) - { - returnA.model>B.model; - } -   - BOOLCmp3 (E a,e b) in { - return(a.golde/a.people) > (b.golde/b.people); to } +   - BOOLCmp4 (E a,e b) the { * return(a.model/a.people) > (b.model/b.people); $ }Panax Notoginseng   - intMain () the { +    intn,m; A     while(cin>>n>>M) the    { +        for(intI=0; i<n;i++) -  { $buf[i].num=i; $Cin>>buf[i].golde>>buf[i].model>>buf[i].people; -  } -   for(intI=0; i<m;i++) the  { -    inttemp;WuyiCin>>temp; thebuf[i]=Buf[temp]; -  } Wu         -Sort (BUF,BUF+M,CMP1);//1 About   for(intI=0; i<m;i++) $  { -  if(i!=0&&buf[i-1].golde==Buf[i].golde) -buf[i].oder[0]=buf[i-1].oder[0]; -  Else Abuf[i].oder[0]=i+1; +   the  } -   $   theSort (BUF,BUF+M,CMP2);//2 the   for(intI=0; i<m;i++) the  { the  if(i!=0&&buf[i-1].model==Buf[i].model) -buf[i].oder[1]=buf[i-1].oder[1]; in  Else thebuf[i].oder[1]=i+1; the   About  } the   the   theSort (Buf,buf+m,cmp3);//3 +   for(intI=0; i<m;i++) -  { the  if(i!=0&& (buf[i-1].golde/buf[i-1].people) = = (buf[i].golde/buf[i].people))Bayibuf[i].oder[2]=buf[i-1].oder[2]; the  Else thebuf[i].oder[2]=i+1; -  } -   the   theSort (buf,buf+m,cmp4);//4 the   for(intI=0; i<m;i++) the  { -  if(i!=0&& (buf[i-1].model/buf[i-1].people) = = (buf[i].model/buf[i].people)) thebuf[i].oder[3]=buf[i-1].oder[3]; the  Else thebuf[i].oder[3]=i+1;94  } the   theSort (buf,buf+m,cmp0); the   for(intI=0; i<m;i++)98  { About  intindex=0; -      for(intj=0;j<4; j + +)101 {102 if(buf[i].oder[j]<Buf[i].oder[index])103index=J;104 } the  106cout<<buf[i].oder[index]<<":"<<index+1<<Endl;107  }108cout<<Endl;109    } the    return 0;111 }   the  

Olympic ranking problems score: 2