Source: PAT A1075 PAT Judge (25 分) Description: The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contai…

1075 PAT Judge (25 分) The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each…

A1075 PAT Judge (25)(25 分) The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For…

题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/677 5-15 PAT Judge   (25分) The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Spe…

1075 PAT Judge (25分)   The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For eac…

原题连接:https://pta.patest.cn/pta/test/16/exam/4/question/677 题目如下: The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each…

用了冒泡和插入排序 果然没有什么本质区别..都是运行超时 用库函数sort也超时 The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one…

题目地址:http://pat.zju.edu.cn/contests/pat-a-practise/1075 此题主要考察细节的处理,和对于题目要求的正确理解,另外就是相同的总分相同的排名的处理一定要熟练,还有就是编译没有通过为零分,没有提交显示为"-": #include <cstdio> #include <vector> #include <algorithm> using namespace std; ; vector<); int…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805393241260032 题意: 有m次OJ提交记录,总共有k道题,n个人.每道题有一个最高分. 现在要统计用户的排名,如果总分相同,完整AC的题目数高的排前面,都一样id小的排前面. 如果没有提交记录,或者提交记录都是-1的用户,就不输出. 思路: 根据题意模拟.PAT的题目都要耐心好好读题啊,各种情况都要看清楚. #include<cstdio> #inc…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

排序题 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <map> using namespace std; const int N = 100005; int score[10]; struct Node { int id; int problem; int get; }person[N]; struct ANS {…

//终于A了,不难却觉着坑多的的题,注意-1的处理,感觉我是受memset置0的束缚了,可以把初试成绩置-1.就不用debug怎么久,注意对于-1的处理,不然漏洞百出 #include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int maxn=10001;struct node{ int id; int r; int submit…

/* * 1.主要就用了个sort对结构体的三级排序 */ #include "iostream" #include "algorithm" using namespace std; ]; struct Node { int id; ] = {-,-,-,-,-,-}; /* 记录每一题的分数 初始化为-2代表没答题 */ ; /* 记录总分 */ ; /* 记录得满分的题目总数 */ bool flag = false; /* 判断该用户能否上ranklist 默…

简单模拟题. 注意一点:如果一个人所有提交的代码都没编译通过,那么这个人不计排名. 如果一个人提交过的代码中有编译不通过的,也有通过的,那么那份编译不通过的记为0分. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<str…

相当于是模拟OJ评测,这里注意最后输出:1.那些所有提交结果都是-1的(即均未通过编译器的),或者从没有一次提交过的用户,不需要输出.2.提交结果为-1的题目,最后输出分数是03.某个题目从没有提交过的,输出'-' #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> #include <queue>…

其中在排名输出上参照了 http://blog.csdn.net/xyzchenzd/article/details/27074665,原先自己写的很繁琐,而且还有一个测试点不通过. #include <iostream> #include <vector> #include <string.h> #include <algorithm> #include <stdio.h> using namespace std; struct PATInfo…

题意: 输入三个正整数N,K,M(N<=10000,K<=5,M<=100000),接着输入一行K个正整数表示该题满分,接着输入M行数据,每行包括学生的ID(五位整数1~N),题号和该题得分(-1表示没通过编译).输出排名,学生ID,总分和每一题的得分,第一优先为总分降序,第二优先为题目AC数降序,第三优先为学生ID升序(提交但未通过编译得分为0,未提交得分为-,不输出没有提交或者提交全都未通过编译的学生信息). trick: 测试点4为有学生先交了得到分的程序后该题后来又交了未通过编译…

2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分子. // 1063 Set Similarity #include <set> #include <map> #include <cstdio> #include <iostream> #include <algorithm> using name…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…