1141. PAT Ranking of Institutions (25)

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分类专栏： PAT甲级（二刷）

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该博客介绍了PAT测试中心根据学生表现对机构进行排名的过程。博主分享了一道编程题的输入输出规格，并提到在考试中因代码细节导致超时的问题，讨论了对double类型处理的不一致性和对排名算法的理解。

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1141. PAT Ranking of Institutions (25)

时间限制

500 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

After each PAT, the PAT Center will announce the ranking of institutions based on their students' performances. Now you are asked to generate the ranklist.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=10⁵), which is the number of testees. Then N lines follow, each gives the information of a testee in the following format:

ID Score School

where "ID" is a string of 6 characters with the first one representing the test level: "B" stands for the basic level, "A" the advanced level and "T" the top level; "Score" is an integer in [0, 100]; and "School" is the institution code which is a string of no more than 6 English letters (case insensitive). Note: it is guaranteed that "ID" is unique for each testee.

Output Specification:

For each case, first print in a line the total number of institutions. Then output the ranklist of institutions in nondecreasing order of their ranks in the following format:

Rank School TWS Ns

where "Rank" is the rank (start from 1) of the institution; "School" is the institution code (all in lower case); "TWS" is the total weighted score which is defined to be the integer part of "Score_B/1.5 + Score_A + Score_T*1.5", where "Score_X" is the total score of the testees belong to this institution on level X; and "Ns" is the total number of testees who belong to this institution.

The institutions are ranked according to their TWS. If there is a tie, the institutions are supposed to have the same rank, and they shall be printed in ascending order of Ns. If there is still a tie, they shall be printed in alphabetical order of their codes.

Sample Input:

10
A57908 85 Au
B57908 54 LanX
A37487 60 au
T28374 67 CMU
T32486 24 hypu
A66734 92 cmu
B76378 71 AU
A47780 45 lanx
A72809 100 pku
A03274 45 hypu

Sample Output:

5
1 cmu 192 2
1 au 192 3
3 pku 100 1
4 hypu 81 2
4 lanx 81 2

这题考试的时候只提交了1次，最后一点超时,以为是map查找或者排序的原因，结果今天把cout改成printf就过了...可惜了3分

我就想说，出题人恶不恶心？？？有的时候对double类型的四舍五入，有的时候向下取整，有意思吗？？？

#include<stdio.h>
#include<string>
#include<math.h>
#include<iostream>
#include<map>
#include<algorithm>
using namespace std;
int person;
map<string,int>mp;
struct node{
	int rank;
	string name;
	double sumscore;
	int stunum;
	node(){
		sumscore=0;
		stunum=0;
	}
}stu[100010];
void change(string& s){//改成小写 
	int i;
	for(i=0;i<s.length();i++){
		if(s[i]>='A'&&s[i]<='Z'){
			s[i]=s[i]+32;
		}
	}
}
int getindex(string s){
	if(mp.find(s)==mp.end()){
		mp[s]=person;
		return person++;
	}
	else{
		return mp[s];
	}
}
bool cmp(node a,node b){
	if((a.sumscore!=b.sumscore))return a.sumscore>b.sumscore;
	else if(a.stunum!=b.stunum)return a.stunum<b.stunum;
	else return a.name<b.name;
}
int main(){
	int n,i,score,index;
	string id,school,tmp;
	scanf("%d",&n);
	for(i=0;i<n;i++){
		cin>>id>>score>>school;
		change(school);//改成小写 
		index=getindex(school);
		stu[index].name=school;
		stu[index].stunum++;
		if(id[0]=='T'){
			stu[index].sumscore=stu[index].sumscore+score*1.5;
		}
		else if(id[0]=='A'){
			stu[index].sumscore=stu[index].sumscore+score;
		}
		else if(id[0]=='B'){
			stu[index].sumscore=stu[index].sumscore+score/1.5;
		}
	}
	for(i=0;i<person;i++){//恶不恶心，有时候四舍五入，有时候向下取整 
		stu[i].sumscore=int(stu[i].sumscore);
	}
	sort(stu,stu+person,cmp);
	stu[0].rank=1;
	for(i=1;i<person;i++){
		stu[i].rank=i+1;
		if(stu[i].sumscore==stu[i-1].sumscore){
			stu[i].rank=stu[i-1].rank;
		}
	}
	printf("%d\n",person);
	for(i=0;i<person;i++){//卧槽.... 
		printf("%d %s ",stu[i].rank,stu[i].name.c_str());
		printf("%.0lf %d\n",stu[i].sumscore,stu[i].stunum);
	}
//	for(i=0;i<person;i++){//最后一点超时.... 
//		cout<<stu[i].rank<<" "<<stu[i].name<<" ";
//		printf("%.0lf ",stu[i].sumscore);
//		cout<<stu[i].stunum<<endl;
//	}
}