浙大 pat 1047题解】的更多相关文章

1047. Student List for Course (25) 时间限制 400 ms 内存限制 64000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Zhejiang University has 40000 students and provides 2500 courses. Now given the registered course list of each student, you are supposed to output…
1035. Password (20) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is…
1025. PAT Ranking (25) 时间限制 200 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneous…
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their…
#include <stdio.h> #include <iostream> #include <algorithm> #include <math.h> #include <string.h> #include <queue> #include <stack> #define N 400 #define ll int using namespace std; struct node{  ll gold,all,pre,n…
1012. The Best Rank (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematic…
1003. Emergency (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount…
1038. Recover the Smallest Number (30) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given {32, 321, 3214, 0229, 87},…
1054. The Dominant Color (20) 时间限制 100 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard Behind the scenes in the computer's memory, color is always talked about as a series of 24 bits of information for each pixel. In an image, the color with the larges…
1059. Prime Factors (25) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km. Input Specificatio…
1023. Have Fun with Numbers (20) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913…
1024. Palindromic Number (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number.…
1061. Dating (20) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Sherlock Holmes received a note with some strange strings: "Let's date! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm". It took him only a minut…
1031. Hello World for U (20) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as: h…
1029. Median (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median…
1048. Find Coins (25) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept a…
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For examp…
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…
一开始是建立了course[2501][40001]数组,存储每节课的学生编号然后for循环两层输出,但这样复杂度为O(2500*40000),也很明显导致最后时间超时后来发现最多40000学生,每个学生最多选20门课,那么总共也就40000*20所以直接就存储学生-课程的信息,然后排个序,按照课程从小到大,课程一样的话则按字典序然后从头扫一遍即可,复杂度O(80000)不过要注意一点,有些课可能并没有出现,所以要做个判断,输出x 0. #include <iostream> #include…
1001. 害死人不偿命的(3n+1)猜想 1002. 写出这个数 1003. 我要通过! 1004. 成绩排名 1005. 继续(3n+1)猜想 1006. 换个格式输出整数 1007. 素数对猜想 1008. 数组元素循环右移问题 1009. 说反话 1010. 一元多项式求导 1011. A+B和C 1012. 数字分类 1013. 数素数 1014. 福尔摩斯的约会 1015. 德才论 1016. 部分A+B 1017. A除以B 1018. 锤子剪刀布 1019. 数字黑洞 1020.…
1047 Student List for Course (25 分) Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses. Input Specification: E…
1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…
pat链接:http://pat.zju.edu.cn 1 #include<stdio.h> 2 int main(){ 3 int a,b; 4 int c; 5 while(scanf("%d %d",&a,&b)!=EOF){ 6 c=a+b; 7 if(c<0){ 8 c=-c; 9 printf("-"); 10 } 11 if(c>=1000000) 12 printf("%d,%03d,%03d\n&…
1036. Boys vs Girls (25) This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students. Input Specification: Each input file contains one test case. Each case contain…
1049. Counting Ones (30) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Inpu…
1047 编程团体赛(20)(20 分) 编程团体赛的规则为:每个参赛队由若干队员组成:所有队员独立比赛:参赛队的成绩为所有队员的成绩和:成绩最高的队获胜. 现给定所有队员的比赛成绩,请你编写程序找出冠军队. 输入格式: 输入第一行给出一个正整数N(<=10000),即所有参赛队员总数.随后N行,每行给出一位队员的成绩,格式为:"队伍编号-队员编号 成绩",其中"队伍编号"为1到1000的正整数,"队员编号"为1到10的正整数,"…
编程团体赛的规则为:每个参赛队由若干队员组成:所有队员独立比赛:参赛队的成绩为所有队员的成绩和:成绩最高的队获胜. 现给定所有队员的比赛成绩,请你编写程序找出冠军队. 输入格式: 输入第一行给出一个正整数N(<=10000),即所有参赛队员总数.随后N行,每行给出一位队员的成绩,格式为:“队伍编号-队员编号 成绩”,其中“队伍编号”为1到1000的正整数,“队员编号”为1到10的正整数,“成绩”为0到100的整数. 输出格式: 在一行中输出冠军队的编号和总成绩,其间以一个空格分隔.注意:题目保证…
早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1001. A+B Format (20) 题意 格式化输出两数之和. 分析 理清输出逻辑即可. 1002. A+B for Polynomials (25) 题意 给定两多项式,相加并格式化输出结果. 分析 两种思路 1.采用链表的处理方式: 2.预设好 int[1005]的数组,…
早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1011. World Cup Betting (20) 题意 给定一个 3*3 的矩阵,找到每行的最大值,格式化输出一个运算结果. 分析 非常简单的模拟题. 1012. The Best Rank (25) 题意 给定学生的学号和三个科目的分数,查询输出对应学生单门科目排名和总分排…
早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1021. Deepest Root (25) 题意 无环连通图也可以视为一棵树,选定图中任意一点作为根,如果这时候整个树的深度最大,则称其为 deepest root. 给定一个图,按升序输出所有 deepest root.如果给定的图有多个连通分量,则输出连通分量的数量. 分析…