1056. Mice and Rice (25) 时间限制 30 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The…

1056. Mice and Rice (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. Th…

1056. Mice and Rice (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. Th…

1056 Mice and Rice (25 分)   Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order…

时间限制 30 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to…

题目 Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse. Fir…

1056. Mice and Rice Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to becom…

1.27刷题2 Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse…

有n个老鼠,第一行给出n个老鼠的重量,第二行给出他们的顺序.1.每一轮分成若干组,每组m个老鼠,不能整除的多余的作为最后一组.2.每组重量最大的进入下一轮.让你给出每只老鼠最后的排名.很简单,用两个数组模拟一下即可order1存储进入当前一轮老鼠的索引顺序order2存储进入下一轮老鼠的索引顺序 如果当前有groups个组,那么会有groups个老鼠进入下一轮,则没有进入下一轮的排名都为groups+1如果只有一个组,那么最大的那个排名即为1. #include <iostream> #inc…

简单模拟. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<stack> #include<vector> using namespace std…

题意: 输入两个正整数N和M(<=1000),接着输入两行,每行N个数,第一行为每只老鼠的重量,第二行为每只老鼠出战的顺序.输出它们的名次.(按照出战顺序每M只老鼠分为一组,剩余不足M只为一组,每组只能有一个胜者,其他老鼠排名均为这一轮胜者数量+1) AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; int n,m; ],b[]; int num; ];…

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse. First…

1056 Mice and Rice(25 分) Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to…

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse. First…

Mice and Rice(queue的用法) Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to b…

题意:略 思路:利用queue来模拟一轮一轮的比赛.自己第一遍做的时候完全没有用queue做的意识,代码写的贼烦最后还只得了17分,非常郁闷.通过本题反映出对queue的应用场景季度不熟悉,STL里面用的最少的就是队列了.另外还有一点,在当前这一轮被淘汰的老鼠排名均为当前组数+1,这一点我也没看出来,自己做的时候拐了18个弯去实现这一点,真是惭愧! 代码: #include <cstdio> #include <queue> using namespace std; struct…

#include <cstdio> #include <climits> #include <cstdlib> #include <vector> #include <list> using namespace std; list<int>::iterator group_pick(list<int> &player, list<int>::iterator &cur, int group_si…

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…

浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1051. Pop Sequence (25) 题意 给定 stack 的容量,给定数据的入栈顺序:从 1 开始的正整数序列,在允许随机的出栈操作的情况下,要求判断某出栈序列是否可能. 比如,告知 stack 容量为 5,入栈序列的最大值为 7.有两个序列需要判断合理性: {1 2 3 4 5 6 7}: 这个序列是可能的,只需每次入栈时都做出栈操作. {3 2 1 7 5 6 4}: 这个序列是不可能的,其中前半部分 3 2 1 是合法的,…

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"…

1051 Pop Sequence (25分) [stack] 简答的栈模拟题,只要把过程想清楚就能做出来. 扫描到某个元素时候,假如比栈顶元素还大,说明包括其本身的在内的数字都应该入栈.将栈顶元素和序列比对即可,相同则弹栈,继续扫描:否则无法生成满足条件的序列.注意栈满时不能入栈 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn =…

考虑到PAT甲级考试和开学后的XCPC比赛,决定寒假把PAT (Advanced Level) Practice刷完,进度条会在这篇博客下更新.由于主要以记录为主,大体上不会像单篇题解那么详细,但是对问题的思考,代码的简洁性.可读性还是有保障的,欢迎看到的小伙伴和我讨论 2021.1.10 1001 A+B Format (20分) 很久没写题了,没想到卡了半个小时,惭愧.这里是要把结果用逗号分隔成三组,即以千为单位,不足的话则不必要填逗号,我最多只添了一个逗号,要看清题目意思再动笔 #incl…

这次,还是把基本的基于我目前最新的Linux源码进行移植到OK6410吧,同时也写下我移植过程中遇到的问题及解决方法,不过有些方法是借鉴网上的,有些是自己加的,会有一些小bug. 一.基本工作 1. 源码下载    https://www.kernel.org/   ,最好是下载stable版本,否则会有小bug.(我现在调试的是stable版本, linux-3.14.4) 2. 拷到自己的文件夹下解压,我下的是.xz后缀的,这样文件比较小,只是解压时多一个步骤. $ xz –d linux-…

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…

I Love my family   12 years  ago, I was born in a happy family, there was a gentle father, a beautiful mother and a handsome brother, all the family were so happy for the birth of me. My family was full of warmth and happiness.   My parents and broth…

好多同学跑来问,用spss的时候使用多重插补的数据集,怎么选怎么用?是不是简单的选一个做分析?今天写写这个问题. 什么时候用多重插补 首先回顾下三种缺失机制或者叫缺失类型: 上面的内容之前写过,这儿就不给大家翻译了,完全随机缺失,缺失量较小的情况下你直接扔掉或者任你怎么插补都可以,影响不大的.随机缺失可以用多重插补很好地处理:非随机缺失,任何方法都没得救的,主分析做完之后自觉做敏感性分析才是正道:这个我好像在之前的文章中给大家解释过原因. When it is plausible that da…

By reducing the activity of one type of gene, scientists said they increased the average life span of mice by about 20%, a feat that in human terms is akin to extending life by about 15 years. manipulation:操作,处理 akin to:同类,近似 Moreover, the researcher…