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 <…

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…

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…

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-];…

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…

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 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…

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…

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…

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…

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…

Source: PAT A1093 Count PAT's (25 分) Description: 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…

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…

1040. 有几个PAT(25) 时间限制 120 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CAO, Peng 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105,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…

1040. 有几个PAT(25)     http://www.patest.cn/contests/pat-b-practise/1040 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105,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…

1025. PAT Ranking (25) 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 Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneous…

A1075 PAT Judge (25)(25 分) The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT. Input Specification: Each input file contains one test case. For…

A1025 PAT Ranking (25)(25 分) 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 immedia…

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…

1040. 有几个PAT(25) 时间限制 120 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CAO, Peng 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105,只包含P.A.T三种字母. 输出格式: 在一行中输出给…

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…

题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1025 题目描述: 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…

题目如下: 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. No…