【一只蒟蒻的刷题历程】 【PAT (Advanced Level) Practice】 1025 PAT排名
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:2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85
9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4
编程能力测试（PAT）由浙江大学计算机科学与技术学院组织。每个测试应该在多个地方同时运行，并且等级列表将在测试后立即合并。现在，您的工作就是编写一个程序来正确合并所有排名列表并生成最终排名。
输入规格：
每个输入文件包含一个测试用例。对于每种情况，第一行都包含一个正数N（≤100），即测试位置的数量。然后是N个等级列表，每个等级列表都包含一个正整数K（≤300），受测试者人数的行，然后是包含注册编号（13位数字）和每个受测试者总分的K行。一行中的所有数字都用空格分隔。
输出规格：
对于每个测试用例，首先在一行中打印被测试者的总数。然后以以下格式打印最终排名列表：
registration_number final_rank location_number local_rank
位置的编号从1到N。输出必须按最终等级的降序排列。具有相同分数的被测者必须具有相同的等级，并且输出必须以其注册号的降序排列。
样本输入：2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85
9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4
结构体排序
直接这样不知道为什么就是错的，非得把第一个数先赋值才可以
for(int i=0;i=0) stu[i].final=stu[i-1].final;
else stu[i].final = i+1;
//c++;
}
代码：
#include
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