PAT甲级1008水题飘过

题目分析:上去下来到达的时间和数量 #include<iostream> using namespace std; ]; int main(){ int n; while(scanf("%d", &n) != EOF){ ; ; ans += n*; ; i <= n; i++){ scanf("%d", &k[i]); if(k[i] > floor){ ans += (k[i] - floor) * ; floor = k…

PAT甲级1019水题飘过

题目分析: 将n转成对应大小的b进制数之后判断是否为回文串,是则Yes,否则No #include<iostream> using namespace std; ]; //存放从0开始b进制的每一位数 void run(int n, int b){ ; ){ k[cnt] = n % b; n /= b; cnt++; } ; ; i < cnt / ; i++){ ]){ flag = ; break; } } ) printf("Yes\n"); else pri…

PAT甲级1006水题飘过

题目分析:由于不存在相同的两个时间(24:00:00和00:00:00不会同时存在),则我们假设两个全局变量存放到达的最早的时间和达到的最晚的时间,设置最早的初值为“23:59:59”,设置最晚的初值为“00:00:00”,只要一个人到达的时间比最早的早则更新最早时间同时将id记录下来,最晚的情况也是同样的 #include<iostream> #include<string> using namespace std; int main(){ int n; while(scanf(…

PAT甲级1009水题飘过

题目分析:简单的多项式的模拟乘法,你可以假设未知数为x,exp为x的指数,coe为x的系数,则很容易就把答案推算出来,注意答案是从指数的高往低输出,同时要注意的是这是多项式的乘法,虽然指数的范围只有0~1000,但是乘法结果的指数范围为0~2000,注意数组定义的范围,开小了会产生段错误(下标越界,访问了不该访问的内存) #include<iostream> using namespace std; struct Node{ int exp; //指数 double coe; //系数 }k[…

PAT甲级1011水题飘过

题目分析:对于输入的数据分三条,选出每条中最大值记录下来,按照题目要求算出最大可能的获利即可 #include<iostream> using namespace std; ]; //k数组存放每次的三个倍率 ]; //存放三次选择的下标 ] = {'W', 'T', 'L'}; //存放输出的三个字符 int main(){ ], &k[], &k[]) != EOF){ double out = 1.0; double ma = 0.0; ; ; i < ; i++)…

PAT甲级1005水题飘过

题目分析:用一个字符串输入之后遍历每一位求和后,不断%10获取最后一位存储下来,逆序用对应的英文单词输出(注意输入为0的情况) #include<iostream> #include<string> using namespace std; ] = {"zero", "one", "two", "three", "four", "five", "six&…

PAT甲级1002水题飘过

#include<iostream> #include<string.h> using namespace std; ]; int main(){ int n1, n2; while(scanf("%d", &n1) != EOF){ memset(a, , sizeof(a)); ; i <= n1; i++){ int x; double y; scanf("%d%lf", &x, &y); a[x] +=…

PAT甲级1001水题飘过

#include<iostream> using namespace std; int main(){ int a, b; while(scanf("%d%d", &a, &b) != EOF){ int c = a + b; ) printf("0\n"); else{ ){ printf("-"); c = -c; } ]; ; ){ ; c /= ; k[cnt++] = x; } ; i >= ; i--…

PAT 甲级 1008 Elevator (20)（20 分）模拟水题

题目翻译: 1008.电梯在我们的城市里,最高的建筑物里只有一部电梯.有一份由N个正数组成的请求列表.这些数表示电梯将会以规定的顺序在哪些楼层停下.电梯升高一层需要6秒,下降一层需要4秒.每次停下电梯将花费5秒. 给你一个请求列表,你需要计算出完成列表里的请求总共花费的时间.一开始电梯在0层.当请求全部完成时,电梯不需要回到底层. 输入说明: 每个输入文件包含一个测试实例.每个实例包含一个正整数N,后面跟着N个数字.所有输入的数字小于100. 输出说明: 对于每个测试实例,在一行中输出总时间.…

PAT甲级训练刷题代码记录

刷题链接:https://www.patest.cn/contests/pat-a-practise 1001 #include <iostream> #include <stdio.h> using namespace std; ]; int main() { int a,b,sum,index; while(scanf("%d%d",&a,&b)!=EOF){ sum=a+b; ){ printf("-"); sum=-s…

HDU 4662 MU Puzzle (2013多校6 1008 水题)

MU Puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 134 Accepted Submission(s): 71 Problem Description Suppose there are the symbols M, I, and U which can be combined to produce strings…

HDU 4639 Hehe （2013多校4 1008 水题）

Hehe Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 33 Accepted Submission(s): 22 Problem Description As we all know, Fat Brother likes MeiZi every much, he always find some topic to talk wi…

PAT甲级 Dijkstra 相关题_C++题解

Dijkstra PAT (Advanced Level) Practice Dijkstra 相关题目录 <算法笔记>重点摘要 1003 Emergency (25) <算法笔记> 10.4.1 Dijkstra 重点摘要对任意给出的图 G(V,E) 和起点 S,终点 T,求 S 到 T 的最短路径 1. 简介解决单源最短路问题只能处理所有边权均非负的情况若出现负数,最好使用 SPFA 算法 2. 邻接矩阵 const int MAXV = 1000; const i…

PAT甲级散列题_C++题解

散列 PAT (Advanced Level) Practice 散列题目录 <算法笔记> 重点摘要 1002 A+B for Polynomials (25) 1009 Product of Polynomials (25) 1084 Broken Keyboard (20) 1092 To Buy or Not to Buy (20) 1116 Come on! Let's C (20) 1121 Damn Single (25) <算法笔记> 4.2 散列重点摘要 1.…

PAT甲级二叉树相关题_C++题解

二叉树 PAT (Advanced Level) Practice 二叉树相关题目录 <算法笔记> 重点摘要 1020 Tree Traversals (25) 1086 Tree Traversals Again (25) 1102 Invert a Binary Tree (25) 1119 Pre- and Post-order Traversals (30) 1127 ZigZagging on a Tree (30) 1138 Postorder Traversal (25) 1…

PAT甲级二叉查找树相关题_C++题解

二叉查找树 PAT (Advanced Level) Practice 二叉查找树相关题目录 <算法笔记> 重点摘要 1099 Build A Binary Search Tree (30) 1115 Counting Nodes in a BST (30) 1143 Lowest Common Ancestor (30) <算法笔记> 9.4 二叉查找树重点摘要二叉查找树静态实现 ⭐ (1) 定义 struct Node{ typename data; int level…

PAT 甲级 1008 Elevator (20)（代码）

1008 Elevator (20)(20 分) The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator…

PAT甲级字符串处理题_C++题解

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…

PAT甲级图相关题_C++题解

图 PAT (Advanced Level) Practice 用到图的存储方式,但没有用到图的算法的题目目录 1122 Hamiltonian Cycle (25) 1126 Eulerian Path (25) 1134 Vertex Cover (25) 1142 Maximal Clique (25) 1154 Vertex Coloring (25) 1122 Hamiltonian Cycle (25) 题目思路用 n != queryV.size() 检查是否 query 覆盖…

PAT 甲级 1008 Elevator

https://pintia.cn/problem-sets/994805342720868352/problems/994805511923286016 The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in spe…

PAT甲级树相关题_C++题解

树目录 <算法笔记>重点摘要 1004 Counting Leaves (30) 1053 Path of Equal Weight (30) 1079 Total Sales of Supply Chain (25) 1090 Highest Price in Supply Chain (25) 1094 The Largest Generation (25) 1106 Lowest Price in Supply Chain (25) <算法笔记> 9.2 树的遍历重点摘要…

PAT甲级堆相关题_C++题解

堆目录 <算法笔记>重点摘要 1147 Heaps (30) 1155 Heap Paths (30) <算法笔记> 9.7 堆重点摘要 1. 定义堆是完全二叉树,树中每个结点的值不小于(或不大于)其左右子结点的值,称之为大顶堆(或小顶堆) const int maxn = 100; int heap[maxn], n = 10; // heap 为堆,n 为元素个数 2. 向下调整 // 对 heap 数组在 [low,high] 范围进行向下调整 // 其中 low 为欲…

2019冬季PAT甲级第四题

#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct node{ int data; node *lchild,*rchild; }tree; ]; tree *build(int l,int r){ if(l>r) return NULL; ,pos=; for(int i=l;i<=r;++i) if(a[i]<mn) mn=a[i],pos=i;…

2019冬季PAT甲级第三题

#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; ]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n,m; cin>>n>>m; ;i<=m;++i){ int x,y; cin>>x>>y; v[x].insert(y);…

PAT甲级——1008 Elevator

PATA1008 Elevator The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one…