1108 Finding Average (20 分) The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A "legal" input is a real number in [-1000, 1000] and is accurate up…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000] and is accurate up to no m…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−] and is accurate up to no more than…

1108 Finding Average (20 分) The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000…

1108. Finding Average (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input number…

1108 Finding Average (20 分) The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000…

Source: PAT A 1108 Finding Average (20 分) Description: The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a…

1108 Finding Average(20 分) The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000]…

简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; ]; int n; int num,num2,pos; bool f() { ; ;s[i]…

求给出数的平均数,当然有些是不符合格式的,要输出该数不是合法的. 这里我写了函数来判断是否符合题目要求的数字,有点麻烦. #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> using namespace std; ; bool islegal(char*str){ int len=strlen(str); ,; ;i<len;i++){ //if…

题意:根据条件判定哪些数是合法的,哪些是不合法的.求其中合法的数的平均值. 思路:字符串处理函数,考虑到最后输出的时候需要控制格式,因此选用scanf()和printf().另外需要了解atof()函数,用的较多的是atoi(),但这里是浮点数的转换,所以不要搞错了.(两者都在头文件<stdlib.h>下) 代码: #include <cstdio> #include <cstring> #include <cstdlib>//atof() #include…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000] and is accurate up to no m…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−] and is accurate up to no more than…

题意: 输入一个正整数N(<=100),接着输入一行N组字符串,表示一个数字,如果这个数字大于1000或者小于1000或者小数点后超过两位或者压根不是数字均为非法,计算合法数字的平均数. trick: 测试点2答案错误原因:题面指出如果只有一个合法数字,输出numbers的时候不加s,即为number. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std;…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000] and is accurate up to no m…

话不多说: 该题要求将给定的所有数分为两类,其中这两类的个数差距最小,且这两类分别的和差距最大. 可以发现,针对第一个要求,个数差距最小,当给定个数为偶数时,二分即差距为0,最小:若给定个数为奇数时,差距为1时最小. 针对第二个要求:则将所有给定的数从小到大排列,将前半部分给那个和小些的集合,其余给那个和大的集合,这样做和差距会最大. 以此思路编写代码如下: #include<cstdio> #include<iostream> #include<algorithm>…

Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性表:栈,队列,链表,顺序表 树:二叉树的建立,二叉树的遍历,完全二叉树,二叉查找树,平衡二叉树,堆,哈夫曼树 图:图的存储和遍历 经典高级算法: 深度优先搜索,广度优点搜索,回溯剪枝 贪心,并查集,哈希映射 最短路径(只考察过单源),拓扑排序(18年9月第一次涉及相关概念,未正式考过),关键路径(未…

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…

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…

字符串处理题 目录 <算法笔记> 重点摘要 1001 A+B Format (20) 1005 Spell It Right (20) 1108 Finding Average (20) 1132 Cut Integer (20) 1140 Look-and-say Sequence <算法笔记> 3.6 字符串处理 重点摘要 注意分析输入输出格式 注意细节和边界情况 1001 A+B Format (20) #include<iostream> using names…

大道至简,知易行难.希望能够坚持刷题. PAT甲级真题题库,附上我的代码. Label Title Score Code Level 1001 A+B Format 20 1001 * 1002 A+B for Polynomials 25 1002 * 1005 Spell It Right 20 1005 * 1008 Elevator 20 1008 * 1009 Product of Polynomials 25 1009 * 1011 World Cup Betting 20 1011…

Average Precision (AP) @[ IoU=0.75 | area= all | maxDets=100 ] = 0.136 Average Precision (AP) @[ IoU=0.50:0.95 | area= small | maxDets=100 ] = 0.000 Average Precision (AP) @[ IoU=0.50:0.95 | area=medium | maxDets=100 ] = 0.198 Average Precision (AP)…

Self-made millionaire Steve Siebold spent 26 years interviewing some of the wealthiest people in the world before condensing his findings in his book "How Rich People Think." He found that the secret to getting rich "is not in the mechanics…

A DNA sequence consists of four letters, A, C, G, and T. The GC-ratio of a DNA sequence is the number of Cs and Gs of the sequence divided by the length of the sequence. GC-ratio is important in gene finding because DNA sequences with relatively high…

Finding Lane Lines on the Road The goals / steps of this project are the following: Make a pipeline that finds lane lines on the road Reflect on your work in a written report Reflection 1. Describe your pipeline. As part of the description, explain h…

Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window. For example,MovingAverage m = new MovingAverage(3);m.next(1) = 1m.next(10) = (1 + 10) / 2m.next(3) = (1 + 10 + 3) / 3m.next(5) = (10 + 3…

转一篇文章,主要是关于VOC中Average Precision指标的 原文出处:https://sanchom.wordpress.com/tag/average-precision/ 还有一篇文章:http://nlp.stanford.edu/IR-book/html/htmledition/evaluation-of-ranked-retrieval-results-1.html…

Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window. For example,MovingAverage m = new MovingAverage(3);m.next(1) = 1m.next(10) = (1 + 10) / 2m.next(3) = (1 + 10 + 3) / 3m.next(5) = (10 + 3…