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…

/* * 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…

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…

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…

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…

PAT甲级:1025 PAT Ranking (25分) 题干 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 imme…

1025 PAT Ranking (25分) 1. 题目 2. 思路 设置结构体, 先对每一个local排序,再整合后排序 3. 注意点 整体排序时注意如果分数相同的情况下还要按照编号排序 4. 代码 #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; struct stu{ int location_number; char…

PTA甲级1094 The Largest Generation (25分) A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population. Input Specificatio…

题目链接:1040 有几个PAT (25 分) 做这道题目,遇到了新的困难.解决之后有了新的收获,甚是欣喜! 刚开始我用三个vector数组存储P A T三个字符出现的位置,然后三层for循环,根据字符次序关系, 统计PAT出现的次数.这样提交后三个测试点超时.代码如下: #include <bits/stdc++.h> using namespace std; ; ]; vector<int> p; vector<int> a; vector<int> t…

题目地址 https://pta.patest.cn/pta/test/15/exam/4/question/713 5-5 堆中的路径   (25分) 将一系列给定数字插入一个初始为空的小顶堆H[].随后对任意给定的下标i,打印从H[i]到根结点的路径. 输入格式: 每组测试第1行包含2个正整数NN和MM(\le 1000≤1000),分别是插入元素的个数.以及需要打印的路径条数.下一行给出区间[-10000, 10000]内的NN个要被插入一个初始为空的小顶堆的整数.最后一行给出MM个下标.…

题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/668 5-6 Root of AVL Tree   (25分) An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they d…

A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…

题意: 给定一次PAT测试的成绩,要求输出考生的编号,总排名,考场编号以及考场排名. 分析: 题意很简单嘛,一开始上来就,一组组输入,一组组排序并记录组内排名,然后再来个总排序并算总排名,结果发现最后一个测试点超时. 发现自己一开始太傻太盲目,其实只要一次性全部输进来,记录好考场编号,一次排序就可以了.既然只排了一次,怎么计算考场排名呢,这里我用了三个数组 ];//记录各个考场当前排到的名次 (当前最后一个人的名次) ];//记录个考场当前排到的最后一个人的分数 ];//记录个考场当前已经排好队…

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>…

题意: 输入一个正整数N(N<=100),表示接下来有N组数据.每组数据先输入一个正整数M(M<=300),表示有300名考生,接下来M行每行输入一个考生的ID和分数,ID由13位整数组成.求输出所有考生的数量并且按照考生在全体中的成绩排名以及ID的字典序输出ID和名词,并标注考生所在的考场编号以及他在考场中的排名. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using names…

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…

PAT 准考证号由 4 部分组成: 第 1 位是级别,即 T 代表顶级:A 代表甲级:B 代表乙级: 第 2~4 位是考场编号,范围从 101 到 999: 第 5~10 位是考试日期,格式为年.月.日顺次各占 2 位: 最后 11~13 位是考生编号,范围从 000 到 999. 现给定一系列考生的准考证号和他们的成绩,请你按照要求输出各种统计信息. 输入格式: 输入首先在一行中给出两个正整数 N(≤)和 M(≤),分别为考生人数和统计要求的个数. 接下来 N 行,每行给出一个考生的准考证号和…

题目 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 i…

题目 原题链接 The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.Now given any string, you are supposed to tell t…

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 {…

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…

A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…

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 i…

哈利·波特要考试了,他需要你的帮助.这门课学的是用魔咒将一种动物变成另一种动物的本事.例如将猫变成老鼠的魔咒是haha,将老鼠变成鱼的魔咒是hehe等等.反方向变化的魔咒就是简单地将原来的魔咒倒过来念,例如ahah可以将老鼠变成猫.另外,如果想把猫变成鱼,可以通过念一个直接魔咒lalala,也可以将猫变老鼠.老鼠变鱼的魔咒连起来念:hahahehe. 现在哈利·波特的手里有一本教材,里面列出了所有的变形魔咒和能变的动物.老师允许他自己带一只动物去考场,要考察他把这只动物变成任意一只指定动物的本事…

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…

字符串 APPAPT 中包含了两个单词 PAT,其中第一个 PAT 是第 2 位(P),第 4 位(A),第 6 位(T):第二个 PAT 是第 3 位(P),第 4 位(A),第 6 位(T). 现给定字符串,问一共可以形成多少个 PAT? 输入格式: 输入只有一行,包含一个字符串,长度不超过1,只包含 P.A.T 三种字母. 输出格式: 在一行中输出给定字符串中包含多少个 PAT.由于结果可能比较大,只输出对 1000000007 取余数的结果. 输入样例: APPAPT 输出样例: 2 #…

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…