题目 原题链接 The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.Now given any string, you are supposed to tell t…

题意: 输入一行由大写字母'P','A','T',组成的字符串,输出一共有多少个三元组"PAT"(相对顺序为PAT即可),答案对1e9+7取模. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ; string s; ],numpa[]; ]; int main(){ ios::sync_with_stdio(false); cin.ti…

http://www.patest.cn/contests/pat-a-practise/1093 The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters. Now g…

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters. Now given any string, you are supposed to tell the numb…

如题,统计PAT出现的个数,注意PAT不一定要相邻,看题目给的例子就知道了. num1代表目前为止P出现的个数,num12代表目前为止PA出现的个数,num123代表目前为止PAT出现的个数. 遇到P,num1++. 遇到A,那么PA的个数为:前面统计的PA的个数(num12)+前面的P与当前A组成的个数(num1) 遇到T,那么PAT的个数为:前面统计的PAT的个数(num123)+前面的PA与当前的T组成的个数(num12) #include <iostream> #include <…

一.技术总结 这是一个逻辑题,题目大职意思是可以组成多少个PAT,可以以A为中心计算两边的P和T,然后数量乘积最后相加便是答案. 还有一个注意的是每次相加后记得mod,取余,不要等到最后加完再取余,会报错可能会溢出. 二.参考代码 #include<iostream> #include<cstring> using namespace std; const int maxn = 100010; const int inf = 1000000007; int leftNump[max…

预处理每个位置之前有多少个P,每个位置之后有多少个T. 对于每个A,贡献的答案是这个A之前的P个数*这个A之后T个数. #include<cstdio> #include<cstring> ; ; long long dp1[maxn],dp2[maxn]; char s[maxn]; int main() { scanf("%s",s); memset(dp1,,sizeof dp1); ]==]=; ;s[i];i++) { dp1[i]=dp1[i-];…

本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/93389073 1093 Count PAT's (25 分)   The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed…

1093. Count PAT's (25) 时间限制 120 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CAO, Peng The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3r…

1093 Count PAT's (25 分) The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters. Now given any string, you are s…

题目链接:1040 有几个PAT (25 分) 做这道题目,遇到了新的困难.解决之后有了新的收获,甚是欣喜! 刚开始我用三个vector数组存储P A T三个字符出现的位置,然后三层for循环,根据字符次序关系, 统计PAT出现的次数.这样提交后三个测试点超时.代码如下: #include <bits/stdc++.h> using namespace std; ; ]; vector<int> p; vector<int> a; vector<int> t…

题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/677 5-15 PAT Judge   (25分) The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Spe…

1075 PAT Judge (25分)   The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For eac…

1025 PAT Ranking (25分) 1. 题目 2. 思路 设置结构体, 先对每一个local排序,再整合后排序 3. 注意点 整体排序时注意如果分数相同的情况下还要按照编号排序 4. 代码 #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; struct stu{ int location_number; char…

PAT甲级:1025 PAT Ranking (25分) 题干 Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged imme…

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters. Now given any string, you are supposed to tell the numb…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

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…

字符串 APPAPT 中包含了两个单词 PAT,其中第一个 PAT 是第 2 位(P),第 4 位(A),第 6 位(T):第二个 PAT 是第 3 位(P),第 4 位(A),第 6 位(T). 现给定字符串,问一共可以形成多少个 PAT? 输入格式: 输入只有一行,包含一个字符串,长度不超过1,只包含 P.A.T 三种字母. 输出格式: 在一行中输出给定字符串中包含多少个 PAT.由于结果可能比较大,只输出对 1000000007 取余数的结果. 输入样例: APPAPT 输出样例: 2 #…

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出两个正整数 N(≤)和 M(≤),分别为考生人数和统计要求的个数. 接下来 N 行,每行给出一个考生的准考证号和…

The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For each case, the first line…

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it i…

题目 Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now i…

题目分析: 由于本题字符串长度有10^5所以直接暴力是不可取的,猜测最后的算法应该是先预处理一下再走一层循环就能得到答案,所以本题的关键就在于这个预处理的过程,由于本题字符串匹配的内容的固定的PAT,所以我们可以这样想,对于一个输入的串,我们找到每个A的位置,只要知道这个A的前面有几个P,这个A的后面有几个T,就可以得到以这个A为中心的所有种数,二者相乘即可,然后如果我们能得到0~s.size()-1范围内每个A的前面有多少个P,每个A后面有多少个T,只要从头遍历一遍并且求和就能得到最终答案,由…

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters. Now given any string, you are supposed to tell the numb…

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…

字符串 APPAPT 中包含了两个单词 PAT,其中第一个 PAT 是第 2 位(P),第 4 位(A),第 6 位(T):第二个 PAT 是第 3 位(P),第 4 位(A),第 6 位(T). 现给定字符串,问一共可以形成多少个 PAT? 输入格式: 输入只有一行,包含一个字符串,长度不超过1,只包含 P.A.T 三种字母. 输出格式: 在一行中输出给定字符串中包含多少个 PAT.由于结果可能比较大,只输出对 1000000007 取余数的结果. 输入样例: APPAPT 输出样例: 2 #…

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters. Now given any string, you are supposed to tell the numb…

题意: 给定一次PAT测试的成绩,要求输出考生的编号,总排名,考场编号以及考场排名. 分析: 题意很简单嘛,一开始上来就,一组组输入,一组组排序并记录组内排名,然后再来个总排序并算总排名,结果发现最后一个测试点超时. 发现自己一开始太傻太盲目,其实只要一次性全部输进来,记录好考场编号,一次排序就可以了.既然只排了一次,怎么计算考场排名呢,这里我用了三个数组 ];//记录各个考场当前排到的名次 (当前最后一个人的名次) ];//记录个考场当前排到的最后一个人的分数 ];//记录个考场当前已经排好队…