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PAT甲级：1036 Boys vs Girls (25分)

题干

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N lines of student information. Each line contains a student‘s name, gender, ID and grade, separated by a space, where name and ID are strings of no more than 10 characters with no space, gender is either F (female) or M (male), and grade is an integer between 0 and 100. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case, output in 3 lines. The first line gives the name and ID of the female student with the highest grade, and the second line gives that of the male student with the lowest grade. The third line gives the difference grade**F?grade**M. If one such kind of student is missing, output Absent in the corresponding line, and output NA in the third line instead.

Sample Input 1:

3
Joe M Math990112 89
Mike M CS991301 100
Mary F EE990830 95

      
    

Sample Output 1:

Mary EE990830
Joe Math990112
6

      
    

Sample Input 2:

1
Jean M AA980920 60

      
    

Sample Output 2:

Absent
Jean AA980920
NA

思路

上sort函数，直接写一个符合题意的cmp函数，使得数组开头第一个就是女生最高分，最后一个是男生最低分。

然后就很简单了~

至于cmp函数怎么写，先比较性别，男生往后排，女生往前排。性别相同时按成绩从高到低排即可~

第二种方法也挺简单，就搜索一遍数组找到男女生的最大值和最小值也可以，复杂度应该比直接排序要小~

这种就不说了，比较简单，看看这种新奇点的方法能不能给自己带来点启发吧~

code

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
struct stu{
	string name;
	char sex;
	string ID;
	int grade;
};
bool cmp(stu a, stu b){//只需写一个cmp函数即可~
	if(a.sex == b.sex) return a.grade > b.grade;
	return a.sex < b.sex;
}
void P(stu s, char sex){
	if(s.sex == sex) printf("%s %s\n", s.name.c_str(), s.ID.c_str());
	else printf("Absent\n");
}
int main(int argc, char** argv) {
	int n = 0, flag = 0;
	scanf("%d", &n);
	vector<stu> score(n);
	for(int i = 0; i < n; i++) score[i].name.resize(15), score[i].ID.resize(15);
	for(int i = 0; i < n; i++) scanf("%s %c %s %d", &score[i].name[0], &score[i].sex, &score[i].ID[0], &score[i].grade);
	sort(score.begin(), score.end(), cmp);
	if(!(score[0].sex == ‘F‘ && score[score.size() - 1].sex == ‘M‘)) flag = 1;
	P(score[0], ‘F‘);
	P(score[score.size() - 1], ‘M‘);
	if(flag) printf("NA\n");
	else printf("%d\n", score[0].grade - score[score.size() - 1].grade);
	return 0;
}

结果

PAT甲级：1036 Boys vs Girls (25分)

标签：with   put   比较   pos   i++   复杂度   结果   lin   lan   

原文地址：https://www.cnblogs.com/cell-coder/p/12829514.html