pat_1012

源地址：http://www.patest.cn/contests/pat-a-practise/1012

题目不难，主要是有一个地方要注意，比如有三个人，他们的分数是 95 95 90，那么第三个人的排名应该是3而不是2，这个注意了，基本上就能够答对。

结构体排序 + map记录。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<set>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-3
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll long long
int n,m;
#define Mod 1000000007
#define N 510
#define M 1000100
const int size = 10010;
const int mod = 9901;
struct Student{
	char id[20];
	int cscore;
	int mscore;
	int escore;
	int ascore;
}a[2015];
bool cmp1(Student x,Student y){
	return (x.ascore>y.ascore);
}
bool cmp2(Student x,Student y){
	return (x.cscore>y.cscore);
}
bool cmp3(Student x,Student y){
	return (x.mscore>y.mscore);
}
bool cmp4(Student x,Student y){
	return (x.escore>y.escore);
}
map<string,pair<int,char> > mp;
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    while(scanf("%d%d",&n,&m)!=EOF){
    	for(int i=1;i<=n;i++){
    		scanf("%s%d%d%d",a[i].id,&a[i].cscore,&a[i].mscore,&a[i].escore);
    		a[i].ascore = (a[i].cscore+a[i].mscore+a[i].escore)/3;
    	}
    	mp.clear();
    	sort(a+1,a+n+1,cmp1);
    	int id=1;
    	int sc=a[1].ascore;
    	int same=1;
    	for(int i=1;i<=n;i++){
    		if(a[i].ascore != sc){
    			id += same;
    			sc = a[i].ascore;
    			same = 1;
    		}else if(i != 1)
    			same++;
    		mp[a[i].id] = make_pair(id,'A');
    	}
    	sort(a+1,a+n+1,cmp2);
    	id=1;
    	sc = a[1].cscore;
    	same = 1;
    	for(int i=1;i<=n;i++){
    		if(a[i].cscore != sc){
    			id += same;
    			sc = a[i].cscore;
    			same = 1;
    		}else if(i != 1)
    			same++;
    		if(id<mp[a[i].id].first)
    			mp[a[i].id] = make_pair(id,'C');
    	}
    	sort(a+1,a+n+1,cmp3);
    	id=1;
    	sc = a[1].mscore;
    	same = 1;
    	for(int i=1;i<=n;i++){
    		if(a[i].mscore != sc){
    			id += same;
    			sc = a[i].mscore;
    			same = 1;
    		}else if(i != 1)
    			same++;
    		if(id<mp[a[i].id].first)
    			mp[a[i].id] = make_pair(id,'M');
    	}
    	sort(a+1,a+n+1,cmp4);
    	id=1;
    	sc = a[1].escore;
    	same = 1;
    	for(int i=1;i<=n;i++){
    		if(a[i].escore != sc){
    			id += same;
    			sc = a[i].escore;
    			same = 1;
    		}else if(i != 1)
    			same++;
    		if(id<mp[a[i].id].first)
    			mp[a[i].id] = make_pair(id,'E');
    	}
    	char sid[20];
    	for(int i=0;i<m;i++){
    		scanf("%s",sid);
    		if(mp.find(sid) == mp.end())
    			printf("N/A\n");
    		else
    			printf("%d %c\n",mp[sid].first,mp[sid].second);
    	}
    }
return 0;
}