PAT排序题---1015 德才论 (25分)

1015 德才论 (25分)

• 分四个等级的排序，想简化的朋友可以用结构体数组容器存储啦~

#include<iostream>
#include<vector>
#include<ctype.h>
#include<algorithm>
#include<sstream>
#include<string>
#include<cstdio>

using namespace std;

/*此类考生按德才总分从高到低排序*/

/*才分不到但德分到线的一类考生
也按总分排序，但排在第一类考生之后*/

/*德才分均低于H,但是德分不低于才分的考生,
按总分排序，但排在第二类考生之后*/

/*其他达到最低线 L 的考生也按总分排序，但排在第三类考生之后*/

struct Node{
	int num, grade1,grade2,sum;
};

int cnt=0;
vector<Node> stu1;
vector<Node> stu2;
vector<Node> stu3;
vector<Node> stu4;

bool cmp(Node a,Node b){
	if(a.sum!=b.sum){
		return a.sum>b.sum;
	}else if(a.grade1!=b.grade1){
		return a.grade1>b.grade1;
	}else{
		return a.num<b.num;
	}
}
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	int n,L,H; cin>>n>>L>>H;
	for(int i=1;i<=n;i++){
		int num,grade1,grade2;
		cin>>num>>grade1>>grade2;
		if(grade1>=L&&grade2>=L){
			cnt++;int sum=grade1+grade2;
			if(grade1>=H&&grade2>=H){
				stu1.push_back({num,grade1,grade2,sum});
				continue;
			}
			if(grade1>=H&&grade2<H){
				stu2.push_back({num,grade1,grade2,sum});
				continue;
			}
			if(grade1<H&&grade2<H&&grade1>=grade2){
				stu3.push_back({num,grade1,grade2,sum});
				continue;
			}
			stu4.push_back({num,grade1,grade2,sum});
		}
	}
	sort(stu1.begin(),stu1.end(),cmp);
	sort(stu2.begin(),stu2.end(),cmp);
	sort(stu3.begin(),stu3.end(),cmp);
	sort(stu4.begin(),stu4.end(),cmp);
	cout<<cnt<<"
";
	for(int i=0;i<stu1.size();i++){
		cout<<stu1[i].num<<" "<<stu1[i].grade1<<" "<<stu1[i].grade2<<"
";
	}
	for(int i=0;i<stu2.size();i++){
		cout<<stu2[i].num<<" "<<stu2[i].grade1<<" "<<stu2[i].grade2<<"
";
	}
	for(int i=0;i<stu3.size();i++){
		cout<<stu3[i].num<<" "<<stu3[i].grade1<<" "<<stu3[i].grade2<<"
";
	}
	for(int i=0;i<stu4.size();i++){
		cout<<stu4[i].num<<" "<<stu4[i].grade1<<" "<<stu4[i].grade2<<"
";
	}
	return 0;
}