PAT A1062 Talent and Virtue (25 分)/B 1015 德才论 (25 分)

最新推荐文章于 2022-03-04 19:30:59 发布

信不信我是机智狗

最新推荐文章于 2022-03-04 19:30:59 发布

阅读量137

点赞数

分类专栏： PAT 记录文章标签：数据结构算法

本文链接：https://blog.youkuaiyun.com/weixin_48165493/article/details/113809954

版权

PAT 同时被 2 个专栏收录

76 篇文章

订阅专栏

记录

75 篇文章

订阅专栏

PAT A1062 Talent and Virtue (25 分)/B 1015 德才论 (25 分)

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people’s talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a “sage（圣人）”; being less excellent but with one’s virtue outweighs talent can be called a “nobleman（君子）”; being good in neither is a “fool man（愚人）”; yet a fool man is better than a “small man（小人）” who prefers talent than virtue.

Now given the grades of talent and virtue of a group of people, you are supposed to rank them according to Sima Guang’s theory.

Input Specification:
Each input file contains one test case. Each case first gives 3 positive integers in a line: N (≤10^5 ), the total number of people to be ranked; L (≥60), the lower bound of the qualified grades – that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (<100), the higher line of qualification – that is, those with both grades not below this line are considered as the “sages”, and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as the “noblemen”, and are also ranked in non-increasing order according to their total grades, but they are listed after the “sages”. Those with both grades below H, but with virtue not lower than talent are considered as the “fool men”. They are ranked in the same way but after the “noblemen”. The rest of people whose grades both pass the L line are ranked after the “fool men”.

Then N lines follow, each gives the information of a person in the format:

ID_Number Virtue_Grade Talent_Grade
where ID_Number is an 8-digit number, and both grades are integers in [0, 100]. All the numbers are separated by a space.

Output Specification:
The first line of output must give M (≤N), the total number of people that are actually ranked. Then M lines follow, each gives the information of a person in the same format as the input, according to the ranking rules. If there is a tie of the total grade, they must be ranked with respect to their virtue grades in non-increasing order. If there is still a tie, then output in increasing order of their ID’s.

Sample Input:
14 60 80
10000001 64 90
10000002 90 60
10000011 85 80
10000003 85 80
10000004 80 85
10000005 82 77
10000006 83 76
10000007 90 78
10000008 75 79
10000009 59 90
10000010 88 45
10000012 80 100
10000013 90 99
10000014 66 60
Sample Output:
12
10000013 90 99
10000012 80 100
10000003 85 80
10000011 85 80
10000004 80 85
10000007 90 78
10000006 83 76
10000005 82 77
10000002 90 60
10000014 66 60
10000008 75 79
10000001 64 90

B 1015 德才论 (25 分) 中文题目：
宋代史学家司马光在《资治通鉴》中有一段著名的“德才论”：“是故才德全尽谓之圣人，才德兼亡谓之愚人，德胜才谓之君子，才胜德谓之小人。凡取人之术，苟不得圣人，君子而与之，与其得小人，不若得愚人。”

现给出一批考生的德才分数，请根据司马光的理论给出录取排名。

输入格式：
输入第一行给出 3 个正整数，分别为：N（≤10
5
），即考生总数；L（≥60），为录取最低分数线，即德分和才分均不低于 L 的考生才有资格被考虑录取；H（<100），为优先录取线——德分和才分均不低于此线的被定义为“才德全尽”，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属于“德胜才”，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属于“才德兼亡”但尚有“德胜才”者，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。

随后 N 行，每行给出一位考生的信息，包括：准考证号德分才分，其中准考证号为 8 位整数，德才分为区间 [0, 100] 内的整数。数字间以空格分隔。

输出格式：
输出第一行首先给出达到最低分数线的考生人数 M，随后 M 行，每行按照输入格式输出一位考生的信息，考生按输入中说明的规则从高到低排序。当某类考生中有多人总分相同时，按其德分降序排列；若德分也并列，则按准考证号的升序输出。

输入样例：
14 60 80
10000001 64 90
10000002 90 60
10000011 85 80
10000003 85 80
10000004 80 85
10000005 82 77
10000006 83 76
10000007 90 78
10000008 75 79
10000009 59 90
10000010 88 45
10000012 80 100
10000013 90 99
10000014 66 60
输出样例：
12
10000013 90 99
10000012 80 100
10000003 85 80
10000011 85 80
10000004 80 85
10000007 90 78
10000006 83 76
10000005 82 77
10000002 90 60
10000014 66 60
10000008 75 79
10000001 64 90

以下是AC的代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100100;
struct person{
	char id[10];
	int ta,vi,tot;
	int kin;//0,1,2,3分别代表圣人，君子，愚人，小人； 
	bool legit;
}per[maxn];
bool cmp(person a,person b){
	if(a.tot != b.tot) return a.tot>b.tot;
	else if(a.vi != b.vi) return a.vi>b.vi;
	else return strcmp(a.id,b.id)<0;
}
int main(void){
	int total,leg,stan;
	int legal=0;
	scanf("%d%d%d",&total,&leg,&stan);
	for(int i=0;i<total;i++){
		per[i].legit=false;
		scanf("%s%d%d",&per[i].id,&per[i].vi,&per[i].ta);
		per[i].tot=per[i].vi+per[i].ta;
		if(per[i].vi>=leg&&per[i].ta>=leg){
			per[i].legit=true;
			legal++;
			if(per[i].legit){//把人分类； 
				if(per[i].vi>=stan&&per[i].ta>=stan){
					per[i].kin=0;//圣人； 
				}else if(per[i].vi>=stan){
					per[i].kin=1;//君子； 
				}else if(per[i].vi>=per[i].ta){
					per[i].kin=2;//愚人; 
				}else per[i].kin=3;//小人; 	
			}		
		}
	}
	sort(per,per+total,cmp);
	printf("%d\n",legal);
	for(int i=0;i<4;i++){
		for(int j=0;j<total;j++){
			if(per[j].legit&&per[j].kin==i){
				printf("%s %d %d\n",per[j].id,per[j].vi,per[j].ta);
			}
		}
	}
	return 0;
}

思路：
排序题。需要熟悉C++标准库中 sort()函数的用法。
首先使用结构体 person存储每个人的 id, vi（virtue，德得分），ta（talent，才得分），tot（total，总分），kin（kind，哪一类，0,1,2,3分别代表圣人，君子，愚人，小人）,legit（是否合法，不合法的没有分类）。cmp函数中，先比总分，然后比德得分，然后比 id 字典序。
（这里答案是先比类别，类别数字小的在前）排序后按照题目要求顺序依次输出就行。我这里最后相当于遍历了数组4次，如果按照答案先排类别，最后只有一层循环就够了。不过这道题这里影响不大。