1130. Infix Expression (25) 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 Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators.…

1130 Infix Expression (25 分) Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file contains one test case. For e…

1130 Infix Expression (25 分)(找规律.中序遍历) 我是先在CSDN上面发表的这篇文章https://blog.csdn.net/weixin_44385565/article/details/89035813 Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the preced…

https://pintia.cn/problem-sets/994805342720868352/problems/994805347921805312 Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specificati…

Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file contains one test case. For each case, the first line give…

Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file contains one test case. For each case, the first line give…

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/6789828.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~~ 水,中序遍历输出即可注意除根节点.叶子节点外,都需要有括号括起来 #include <iostream> #include <cstdio> #include <algorithm> #include <string>…

题意:给出一个语法树(二叉树),输出相应的中缀表达式. 思路:很显然,通过中序遍历来做.通过观察,发现除了根结点之外的所有非叶结点的两侧都要输出括号,故在中序遍历时判断一下即可. 代码: #include <cstdio> #include <cstring> struct Node{ ]; int left,right; }Tree[]; ;//根结点 void inOrderTraversal(int v) { ){ || Tree[v].right!=-)) printf(&…

1130. Infix Expression (25) 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 Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators.…

Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file contains one test case. For each case, the first line give…

Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file contains one test case. For each case, the first line give…

Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file contains one test case. For each case, the first line give…

Source: PAT A1130 Infix Expression (25 分) Description: Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators. Input Specification: Each input file con…

#include<iostream> #include<stack> #include<string> using namespace std; char compare(char tp, char op) { if (((tp == '+' || tp == '-') && (op == '*' || op == '/')) || (tp == '#')) return '<'; else if (tp == '('&&op !=…

#include<iostream> #include<stack> #include<string> using namespace std; char compare(char tp, char op) { if (((tp == '+' || tp == '-') && (op == '*' || op == '/')) || (tp == '#')) return '<'; return '>'; } int compute(int…

1128 N Queens Puzzle(20 分) 题意:N皇后问题.按列依次给定N个皇后的行号,问N个皇后是否能同时不存在行冲突.列冲突和主副对角线冲突. 分析: 1.根据题意一定不存在列冲突,所以要考虑行冲突和主副对角线冲突.(做题时太天真,只考虑了主副对角线) 2.若某皇后的位置由(x,y)表示,则x+y相同的皇后一定处于同一副对角线:x-y+N相同的皇后一定处于同一主对角线. #include<cstdio> #include<cstring> #include<c…

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…

2019 秋季 PAT (Advanced Level) C++题解 考试拿到了满分但受考场状态和知识水平所限可能方法不够简洁,此处保留记录,仍需多加学习.备考总结(笔记目录)在这里 7-1 Forever (20 分) "Forever number" is a positive integer A with K digits, satisfying the following constrains: the sum of all the digits of A is m; the…

1121 Damn Single 模拟 // 1121 Damn Single #include <map> #include <vector> #include <cstdio> #include <iostream> #include <algorithm> using namespace std; map<int, int> m, vis; vector<int> p; ; int a[N]; int main()…

Example : Infix : (A+B) * (C-D) ) Postfix: AB+CD-* 算法: 1. Scan the infix expression from left to right. 2. If the scanned character is an operand, append it to result. 3. Else 3.1 If the precedence of the scanned operator is greater than the preceden…

Conversion Algorithm 1.操作符栈压入"#": 2.依次读入表达式的每个单词: 3.如果是操作数则压入操作数栈: 4.如果是操作符,则将操作符栈顶元素与要读入的操作符进行优先级比较 (4.1)如果读入的是 ')',则将操作符栈中的元素压入操作数栈直至遇到 '(': (4.2)如果读入的是 '(',压入操作符栈: (4.3)如果栈顶元素优先级低,压入操作符栈: (4.4)如果读入的元素不为'#',以及栈顶元素优先级高,则将栈顶元素压入操作数栈,将读入的元素压入操作符栈…

中缀表达式的计算 利用两个栈来实现,操作数栈,操作符栈 只支持个位数运算 最后必须输入一个'#' #include<iostream> using namespace std; template<typename ElementType> struct Node { ElementType data; Node<ElementType>* next; }; template<typename ElementType> class LinkStack { pu…

Infix : An expression is called the Infix expression if the operator appears in between the operands in the expression. Simply of the form (operand1 operator operand2).Example : (A+B) * (C-D) Prefix : An expression is called the prefix expression if…

Infix expression: The expression of the form a op b. When an operator is in-between every pair of operands.Postfix expression: The expression of the form a b op. When an operator is followed for every pair of operands. Input : abc++ Output : (a + (b…

Infix : An expression is called the Infix expression if the operator appears in between the operands in the expression. Simply of the form (operand1 operator operand2).Example : (A+B) * (C-D) Prefix : An expression is called the prefix expression if…

题目描述 将中缀表达式(infix expression)转换为后缀表达式(postfix expression).假设中缀表达式中的操作数均以单个英文字母表示,且其中只包含左括号'(',右括号‘)’和双目算术操作符+,-,*,/. 输入格式 第一行是测试样例个数n.以下n行,每行是表示中缀表达式的一个字符串(其中只包含操作数和操作符和左右括号,不包含任何其他字符),长度不超过100个字符. 输出格式 为每一个测试样例单独一行输出对应后缀表达式字符串(其中只包含操作数和操作符,不包含任何其他字符…

Known Notation Time Limit: 2 Seconds      Memory Limit: 65536 KB Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathematics and computer science. It is also known as postfix notation since every operator in an expres…

K - Known Notation Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3829 Appoint description:  System Crawler  (2015-08-15) Description Do you know reverse Polish notation (RPN)? It is a known nota…

public class Test {    public static void main(String[] args) {     SimpleCalculator s=new SimpleCalculator();  String methord="80*(1+0.5)"; //test   double d=s.evaluate(methord );   System.out.println(d);  } }         import java.util.Scanner;…