PAT 1036 Boys vs Girls (25)
2017-03-19 21:00
281 查看
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 gradeF-gradeM.
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:
Sample Output 1:
Sample Input 2:
Sample Output 2:
/* 考察结构体排序 QvQ..
* 午睡操场传来蝉的声音 多少年后也还是很好听~~
*/
#include "iostream"
#include "vector"
#include "cstring"
#include "string"
#include "stack"
#include "algorithm"
using namespace std;
struct Node {
char name[11];
char gender;
char id[11];
int grade;
};
bool cmp(const Node &n1, const Node &n2) {
return n1.grade > n2.grade;
}
bool cmp1(const Node &n1, const Node &n2) {
return n1.grade < n2.grade;
}
int main() {
int n;
vector<Node>v1, v2;
cin >> n;
for (int i = 0; i < n; i++) {
Node node;
cin >> node.name >> node.gender >> node.id >> node.grade;
if (node.gender == 'M') {
v1.push_back(node);
}
if (node.gender == 'F') {
v2.push_back(node);
}
}
sort(v1.begin(),v1.end(),cmp1);
sort(v2.begin(), v2.end(), cmp);
if (v2.size()) {
cout << v2[0].name << " " << v2[0].id << endl;
}
else {
cout << "Absent" << endl;
}
if (v1.size()) {
cout << v1[0].name << " " << v1[0].id << endl;
}
else {
cout << "Absent" << endl;
}
if(v1.size()&&v2.size()){
cout << v2[0].grade - v1[0].grade << endl;
}
else {
cout << "NA" << endl;
}
return 0;
}
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 gradeF-gradeM.
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 AA980920NA
/* 考察结构体排序 QvQ..
* 午睡操场传来蝉的声音 多少年后也还是很好听~~
*/
#include "iostream"
#include "vector"
#include "cstring"
#include "string"
#include "stack"
#include "algorithm"
using namespace std;
struct Node {
char name[11];
char gender;
char id[11];
int grade;
};
bool cmp(const Node &n1, const Node &n2) {
return n1.grade > n2.grade;
}
bool cmp1(const Node &n1, const Node &n2) {
return n1.grade < n2.grade;
}
int main() {
int n;
vector<Node>v1, v2;
cin >> n;
for (int i = 0; i < n; i++) {
Node node;
cin >> node.name >> node.gender >> node.id >> node.grade;
if (node.gender == 'M') {
v1.push_back(node);
}
if (node.gender == 'F') {
v2.push_back(node);
}
}
sort(v1.begin(),v1.end(),cmp1);
sort(v2.begin(), v2.end(), cmp);
if (v2.size()) {
cout << v2[0].name << " " << v2[0].id << endl;
}
else {
cout << "Absent" << endl;
}
if (v1.size()) {
cout << v1[0].name << " " << v1[0].id << endl;
}
else {
cout << "Absent" << endl;
}
if(v1.size()&&v2.size()){
cout << v2[0].grade - v1[0].grade << endl;
}
else {
cout << "NA" << endl;
}
return 0;
}
相关文章推荐
- PAT甲级 1036. Boys vs Girls (25)
- PAT-A-1036. Boys vs Girls (25)
- PAT 1036. Boys vs Girls (25)(简单成绩排序)
- 浙江大学PAT_甲级_1036. Boys vs Girls (25)
- PAT甲级1036. Boys vs Girls (25)
- 【C++】PAT(advanced level)1036. Boys vs Girls (25)
- 【PAT甲级】1036. Boys vs Girls (25)
- PAT (Advanced Level) Practise 1036 Boys vs Girls (25)
- PAT 1036. Boys vs Girls (25)
- PAT刷题:1036. Boys vs Girls (25)
- PAT_A 1036. Boys vs Girls (25)
- PAT 甲级 1036. Boys vs Girls (25)
- 【PAT】【Advanced Level】1036. Boys vs Girls (25)
- PAT(甲级)1036. Boys vs Girls (25)
- PAT 1036. Boys vs Girls (25)
- 1036. Boys vs Girls (25)-PAT甲级真题
- pat 1036. Boys vs Girls (25)
- PAT 1036. Boys vs Girls (25)
- 1036. Boys vs Girls (25)【水题】——PAT (Advanced Level) Practise
- PAT (Advanced Level) 1036. Boys vs Girls (25) 结构体排序