Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question asked. Students may write correct answers in different forms which makes grading very ha…

原题目网址:http://poj.org/problem?id=1686 题目中文翻译: Description 数学教师懒得在考卷中给一个问题评分,因为这个问题中,学生会为所问的问题提出一个复杂的公式,但是学生可以用不同的形式写出正确的答案,这使得评分非常困难. 所以,教师需要计算机程序员的帮助,或许你可以提供帮助. 你应该编写一个程序来阅读不同的公式,并确定它们是否在算术上相同.   Input 输入的第一行包含一个整数N(1 <= N <= 20),即测试用例的数量. 在第一行之后,每个…

Problem Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question asked. Students may write correct answers in different forms which makes grading…

题目链接:http://poj.org/problem?id=1686 思路分析:该问题为表达式求值问题,对于字母使用浮点数替换即可,因为输入中的数字只能是单个digit. 代码如下: #include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <string> using namespace…

  Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3721   Accepted: 1290 Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question ask…

因为这个题目说明了优先级的规定,所以可以从左到右直接运算,在处理嵌套括号的时候,可以使用递归的方法,给定每一个括号的左右边界,伪代码如下: int Cal(){ if(括号)  sum += Cal(); else sum += num; return sum; } 但是这个题目着实坑了我一下,见过WA了,没见过TLE呢……我因为没有看到有空格这个条件,无线TLE,又是消除函数又是改用数组模拟栈,其实就是读入出错和忘记了处理空格,改了之后,成功AC了.代码如下: #include<iostrea…

“模拟“题,运用哈希,不断地按照一定运算规律对一个结果进行计算,如果重复出现就停止并且输出该数.注意到仔细看题,这种题一定要细心! POJ - 2183 Bovine Math Geniuses Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Description Farmer John loves to help the cows further their mathematical skills…

题目链接:http://poj.org/problem?id=2389 题目大意: 大数相乘. 解题思路: java BigInteger类解决 o.0 AC Code: import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while (sc.h…

题目链接:http://poj.org/problem?id=3159 题意:给出m给 x 与y的关系.当中y的糖数不能比x的多c个.即y-x <= c  最后求fly[n]最多能比so[1] 多多少糖? 差分约束问题, 就是求1-n的最短路,  队列实现spfa 会超时了,改为栈实现,就可以 有负环时,用栈比队列快 数组开小了,不报RE,报超时 ,我晕 #include <iostream> #include <cstdlib> #include <cstdio>…

Feel Good Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20408   Accepted: 5632 Case Time Limit: 1000MS   Special Judge Description Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated…