简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; ][]={ "tret","jan", "f…

没什么好说的,注意字符串的处理,以及当数字是13的倍数时,只需高位叫法的单词.比如26,是“hel”,而不是“hel tret”. 代码: #include <iostream> #include <cstdio> #include <algorithm> #include <map> #include <string> #include <string.h> using namespace std; ][]={"tret&…

题意: 输入一个正整数N(<100),接着输入N组数据每组包括一行字符串,将其翻译为另一个星球的数字. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ][]={","tam","hel","maa","huh","tou","kes…

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.…

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…

People on Mars count their numbers with base 13: Zero on Earth is called "tret" on Mars. The numbers 1 to 12 on Earch is called "jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec" on Mars, respectively. For the next higher dig…

简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; ]; int n; int num,num2,pos; bool f() { ; ;s[i]…

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)…

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 floor, and 4 seco…

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…