1062. Talent and Virtue (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Li About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a…

题目信息 1062. Talent and Virtue (25) 时间限制200 ms 内存限制65536 kB 代码长度限制16000 B About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding i…

1062 Talent and Virtue (25 分)   About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage…

/* L (>=60), the lower bound of the qualified grades -- that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (<100), the higher line of qualification ,those with both grades not below this line…

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣人)"; being less excellent…

难得的一次ac 题目意思直接,方法就是对virtue talent得分进行判断其归属类型,用0 1 2 3 4 表示 不合格 sage noblemen foolmen foolmen 再对序列进行排序 优先级  类型>total grade>virtue>id #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespac…

简单排序.题意较长. #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<string> #include<stack> #include<map> #include<algorithm> using namespace std; struct X { int id;…

水题,分组排序即可. #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> using namespace std; /* 太水水水... */ +; int cnt1,cnt2,cnt3,cnt4; struct Node{ int id; int tot; int vir; int tal; bool operator<(const Node t…

题意: 输入三个正整数N,L,H(N<=1E5,L>=60,H<100,H>L),分别代表人数,及格线和高水平线.接着输入N行数据,每行包括一个人的ID,道德数值和才能数值.一个人的道德和才能都不低于H时为圣人,道德不低于H才能不低于L时为贵族,道德和才能都不低于L且道德不低于才能时为愚者,道德和才能都不低于L且道德低于才能时为小人(圣人优先判断,圣人的才能可以高于道德).分别在自己的所在的群体内排序,依照道德才能总和降序排序,第二优先为才能的数值降序,第三优先为id的升序.按照题…

Source: PAT A1062 Talent and Virtue (25 分) Description: About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent an…

1062 Talent and Virtue (25 分) About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣…

1062 Talent and Virtue (25 分) About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣…

1. 在排序的过程中,注意边界的处理(小于.小于等于) 2. 对于B-level,这题是比較麻烦一些了. 源代码: #include <cstdio> #include <vector> #include <algorithm> using namespace std; struct People { int m_id; int m_virtue; int m_talent; People(int id, int virtue, int talent): m_id(id…

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣人)"; being less excellent…

#include <cstdio> #include <cstdlib> #include <cstring> #include <vector> #include <algorithm> using namespace std; class Man { public: ]; int talent; int virtue; }; bool mycmp(const Man& a, const Man& b) { int ta = a…

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣人)"; being less excellent…

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣人)"; being less excellent…

技术要点就是,一个是cmp函数的书写,首先应该分清楚排序关系,然后按照顺序依次排下去. 还有这里有一个巧妙点就是,在结构体中加入了类别这个标签. 学会抽象分类解决,排序比较函数cmp本质工作就是比较结构体里面的大小,不应该加入其它. 参考代码: #define _CRT_SECURE_NO_WARNINGS #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> usi…

这四道题来自 13 年 08 月 30 的 PAT 测试. 代码量不大,思路也比较直接.不过第一题的处理逻辑不太清晰,需要好好把握.稍有不慎就掉进坑里了(很多人被这道 20'的题坑了一个多小时心慌意乱我会乱说-,-?). PAT advanced level 全部源码:请戳 1061. Dating (20) 题意 题意比较模糊,需要仔细对照 Sample 的数据理清思路.给定四个字符串,每个不超过 60 个字符,不含空格.要求从中找到符合如下规定的三个字符(或者它们的位置),并转化成一个时间的…

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…

最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题 要求数最短路径有多…

本文为PAT甲级分类汇编系列文章. 排序题,就是以排序算法为主的题.纯排序,用 std::sort 就能解决的那种,20分都算不上,只能放在乙级,甲级的排序题要么是排序的规则复杂,要么是排完序还要做点什么的. 在1051至1100中有6道: 题号 标题 分数 大意 时间 1055 The World's Richest 25 限定范围排序结果 500ms 1056 Mice and Rice 25 分组排序 200ms 1062 Talent and Virtue 25 一定规则的排序 400m…

今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是modern C++. 为什么要整理呢,因为我2019年9月要考PAT甲级,虽然是第一次考,虽然只学了数据结构(https://mooc.study.163.com/course/1000033001?tid=2402970002#/info),但我要冲着高分(2019年9月8日更新:满分)去. 下…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

Tech companies dominate Glassdoor’s ranking of the highest paying companies in the U.S., snagging 20 of the top 25 spots. But no tech company ranks higher than Juniper Networks, which pays its workers a median total compensation of $157,000. The next…

http://highscalability.com/blog/2014/7/21/stackoverflow-update-560m-pageviews-a-month-25-servers-and-i.html The folks at Stack Overflow remain incredibly open about what they are doing and why. So it’s time for another update. What has Stack Overflow…

Cocos2d-X开发教程-捕鱼达人 Cocos2-x development tutorial - fishing talent 作者:韩梦飞沙 Author:han_meng_fei_sha 邮箱:313134555@qq.com E-mail: 313134555 @qq.com 视频目录及链接:Video directory and links: 1 Cocos2D-X介绍  12:29  2 Cocos2D-X游戏安装过程和创建项目  10:14  3 Cocos2D-X启动流程-OC…

程序的载入和运行(五)--<x86汇编语言:从实模式到保护模式>读书笔记25 前面几篇博文最终把代码分析完了.这篇就来说说代码的编译.运行和调试. 1.代码的编译及写入镜像文件 之前我们都是在命令行输入命令进行编译和写入.源文件少的时候还不认为麻烦,当源文件多了,就会认为特别麻烦.有没有简单的方法呢? 当然有,就是用make工具. 1.1.什么是make工具 make是一个命令工具,它解释Makefile中的指令.在Makefile文件里描写叙述了整个project全部文件的编译顺序.编译规则…