1153 Decode Registration Card of PAT】的更多相关文章

1095 解码PAT准考证/1153 Decode Registration Card of PAT(25 分) PAT 准考证号由 4 部分组成: 第 1 位是级别,即 T 代表顶级:A 代表甲级:B 代表乙级: 第 2~4 位是考场编号,范围从 101 到 999: 第 5~10 位是考试日期,格式为年.月.日顺次各占 2 位: 最后 11~13 位是考生编号,范围从 000 到 999. 现给定一系列考生的准考证号和他们的成绩,请你按照要求输出各种统计信息. 输入格式: 输入首先在一行中给…
A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…
A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…
A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…
A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…
Source: PAT A1153 Decode Registration Card of PAT (25 分) Description: A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits ar…
Decode Registration Card of PAT PAT-1153 这里需要注意题目的规模,并不需要一开始就存储好所有的满足题意的信息 这里必须使用unordered_map否则会超时 vector的使用需要注意,只有一开始赋予了容量才能读取. 不需要使用set也可以 #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<cst…
A registration card number of PAT consists of 4 parts: the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic; the 2nd - 4th digits are the test site number, ranged from 101 to 999; the 5th - 10th digits…
1152 Google Recruitment 思路:判断素数 #include<bits/stdc++.h> using namespace std; const int maxn = 1100; int a[maxn]; int n,k; long long getNum(int pos){ long long x = 0; for(int i=pos;i<=pos+k-1;i++){ x = x*10 + a[i]; } return x; } bool prime(long lo…
排序题 PAT (Advanced Level) Practice 排序题 目录 <算法笔记> 6.9.6 sort()用法 <算法笔记> 4.1 排序题步骤 1012 The Best Rank (25) 1083 List Grades (25) 1137 Final Grading (25) 1141 PAT Ranking of Institutions (25) 1153 Decode Registration Card of PAT (25) <算法笔记>…