PAT A1025
来源:互联网 发布:好听的淘宝会员名大全 编辑:程序博客网 时间:2024/06/11 03:58
原题如下{ps.第一次接触PAT甲级}
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive number N (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.
Output Specification:
For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:
registration_number final_rank location_number local_rank
The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.
Sample Input:251234567890001 951234567890005 1001234567890003 951234567890002 771234567890004 8541234567890013 651234567890011 251234567890014 1001234567890012 85Sample Output:
91234567890005 1 1 11234567890014 1 2 11234567890001 3 1 21234567890003 3 1 21234567890004 5 1 41234567890012 5 2 21234567890002 7 1 51234567890013 8 2 31234567890011 9 2 4
代码如下:
其中利用了结构体,排序函数cmp
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct Student{char id[15];int score;int location_number;int local_rank;}stu[30010];bool cmp(Student a,Student b){if(a.score!=b.score) return a.score>b.score;else return strcmp(a.id,b.id)<0;} int main(){int n,k,num=0;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&k);for(int j=0;j<k;j++){scanf("%s %d",stu[num].id,&stu[num].score);stu[num].location_number=i;num++; }printf("%d\t%d",k,num);sort(stu+num-k,stu+num,cmp);stu[num-k].local_rank=1;for(int j=num-k+1;j<num;j++){ //这里不小心把int j=num-k+j,结果出来以后除了每组的第一名有localrank其他都为0if(stu[j].score==stu[j-1].score){stu[j].local_rank=stu[j-1].local_rank;}else{stu[j].local_rank=j+1-(num-k);}}}printf("%d\n",num);sort(stu,stu+num,cmp);int r=1;for(int i=0;i<num;i++){if(i>0&&stu[i].score!=stu[i-1].score){r=i+1;}printf("%s ",stu[i].id);printf("%d %d %d\n",r,stu[i].location_number,stu[i].local_rank);}return 0;}
- PAT A1025
- pat a1025
- PAT-A1025
- PAT A1025
- PAT A1025
- PAT A1025 PAT Ranking
- PAT a1025题解
- A1025. PAT Ranking (25)
- PAT甲级 A1025.PAT RANKING
- 【PAT】A1025. PAT Ranking (25)
- PAT A1025. PAT Ranking (25)
- PAT A1025. PAT Ranking (25)
- PAT-A1025 PAT Ranking(25)
- PAT甲级练习题A1025. PAT Ranking (25)
- PAT:A1025. PAT Ranking (0/25)
- PAT A1025多考场考生排序
- 清澄A1025. 字符串对比
- 刷清橙OJ--A1025.字符串对比
- PAT甲级 1002 A+B for Polynomials (25)
- 算法导论(二)——排序算法整理
- python3.6.2+django1.11+mysql
- 提供创建XML标记和数据结构的简单方法
- 记录zlib在windows上编译使用调研的资料和过程
- PAT A1025
- [caioj1442][主席树][树状数组]第K大的树 II
- python创建文件时去掉非法字符
- 认识struts2
- 深度学习指南
- ECharts开发入门
- java基本教程之join方法详解 java多线程
- 中山大学算法课程题目详解(第六周)
- 机器学习及实践 2.1.1.1 线性分类器