PAT a1083题解
来源:互联网 发布:ios 网络协议 编辑:程序博客网 时间:2024/06/15 21:08
#include <cstdio>#include <algorithm>using namespace std;struct Student{char name[15];char id[15];int grade;}stu[10000];bool cmp(Student a, Student b){return a.grade > b.grade;}int main(){bool hasInInterval = false;int n;scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%s %s %d", stu[i].name, stu[i].id, &stu[i].grade);}int min, max;scanf("%d%d", &min, &max);sort(stu, stu + n, cmp);for(int i = 0; i < n; i++){if(stu[i].grade >= min && stu[i].grade <= max){hasInInterval = true;printf("%s %s\n", stu[i].name, stu[i].id);}}if(!hasInInterval){printf("NONE");}return 0;}
阅读全文
1 0
- PAT a1083题解
- PAT A1083
- PAT-A1083
- PAT A1083. List Grades (25)
- PAT A1083 list grades (25)
- PAT-A1083. List Grades (25)(排序)
- PAT 1001-1010 题解
- PAT 1011-1020 题解
- PAT 1021-1030 题解
- PAT 1031-1040 题解
- PAT 1041-1050 题解
- PAT 1051-1060 题解
- PAT(basic level)题解
- PAT乙级题解
- PAT题解目录
- PAT甲级题解目录
- PAT乙级题解目录
- PAT甲级1001题解
- javascript中私有属性的实现。
- linux下非阻塞的tcp研究
- WebSocket :用WebSocket实现推送你必须考虑的几个问题
- Java基础 String类
- PAT a1082题解
- PAT a1083题解
- PAT a1084题解
- PAT a1086题解
- PAT a1088题解
- PAT a1089题解
- PAT a1090题解
- PAT a1091题解
- PAT a1092题解
- p5.js入门教程(3) 平滑过渡(Easing)