1088 Rational Arithmetic(20 分) For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient. Input Specification: Each input file contains one test case, which gives in o…

For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient. Input Specification: Each input file contains one test case, which gives in one line the two rational number…

For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient. Input Specification: Each input file contains one test case, which gives in one line the two rational number…

题目 For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, diference, product and quotient. Input Specification: Each input file contains one test case, which gives in one line the two rational numb…

输入为两个分数,让你计算+,-,*,\四种结果,并且输出对应的式子,分数要按带分数的格式k a/b输出如果为负数,则带分数两边要有括号如果除数为0,则式子中的结果输出Inf模拟题最好自己动手实现,考验细节处理,其它没啥好说的. #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ]; ]; long long GC…

题意: 给出两个分式(a1/b1 a2/b2),分子.分母的范围为int型,且确保分母不为0.计算两个分数的加减乘除,结果化为最简的形式,即"k a/b",其中若除数为0的话,输出Inf. 思路: 分数四则运算的模板题.分析详见:基础数学问题 代码: #include <stdio.h> #include <string.h> #include <stdlib.h>//abs() typedef long long LL; struct Fracti…

简单题. 注意:读入的分数可能不是最简的.输出时也需要转换成最简. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<iostream> #include<algor…

题意: 输入两个分数(分子分母各为一个整数中间用'/'分隔),输出它们的四则运算表达式.小数需要用"("和")"括起来,分母为0的话输出"Inf"(输入的分母保证不为0). trick: 测试点2很容易溢出,建议遇到乘法时先化简分数并且不要用相乘小于零来判断异号. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespac…

1.注意在数字和string转化过程中,需要考虑数字不是只有一位的,如300转为"300",一开始卡在里这里, 测试用例: 24/8 100/10 24/11 300/11 2.该题用到了欧几里德算法求最小公约数gcd(a,b) 算法如下: //欧几里德算法求最大公约数gcd,其中a>b long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a%b); } AC代码如下: //#include&l…

1088. Rational Arithmetic (20) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient. Input Spe…