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…

简单模拟题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; struct FenShu { long l…

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…

PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (o…

PAT (Advanced Level) Practice 1008 Elevator (20 分) 凌宸1642 题目描述: 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.…

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…

To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase)…

没什么难的,简单模拟题 #include <iostream> using namespace std; int main() { int num; cin>>num; int cost = 0; int curFloor = 0; while (num--) { int floor; cin>>floor; int tmp = floor - curFloor; cost += tmp > 0 ? 6 * tmp : -4 * tmp; cost += 5; c…

找出一定没问题的字符(即一连串的额字符x个数能被k整除的),剩下的字符都是可能有问题的. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; int k;…