PAT_A1093#Count PAT's
Source:
Description:
The string
APPAPTcontains twoPAT'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 number of
PAT's contained in the string.
Input Specification:
Each input file contains one test case. For each case, there is only one line giving a string of no more than 1 characters containing only
P,A, orT.
Output Specification:
For each test case, print in one line the number of
PAT's contained in the string. Since the result may be a huge number, you only have to output the result moded by 1000000007.
Sample Input:
APPAPT
Sample Output:
2
Keys:
- 模拟题
Code:
#include <cstdio>
#include <string>
#include <iostream>
using namespace std;
const int MOD = ; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("test.txt", "r", stdin);
#endif string s;
cin >> s;
long long t=, p=, a=;
for(int i=s.size()-; i>=; i--)
{
if(s[i]=='P') p = (p+a)%MOD;
else if(s[i]=='A') a = (a+t)%MOD;
else t = (t+)%MOD;
}
printf("%lld\n", p); return ;
}
PAT_A1093#Count PAT's的更多相关文章
- PAT1093: Count PAT's
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 strin ...
- PAT 1093 Count PAT's[比较]
1093 Count PAT's (25 分) The string APPAPT contains two PAT's as substrings. The first one is formed ...
- pat1093. Count PAT's (25)
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 strin ...
- PAT甲级——1093 Count PAT's (逻辑类型的题目)
本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/93389073 1093 Count PAT's (25 分) ...
- 1093. Count PAT's (25)
The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and ...
- 1093. Count PAT's
The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and ...
- 1093. Count PAT’s (25)-统计字符串中PAT出现的个数
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 ...
- 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 ...
- PAT (Advanced Level) Practise - 1093. Count PAT's (25)
http://www.patest.cn/contests/pat-a-practise/1093 The string APPAPT contains two PAT's as substrings ...
随机推荐
- PAT_A1077#Kuchiguse
Source: PAT A1077 Kuchiguse (20 分) Description: The Japanese language is notorious for its sentence ...
- apache 安装配置 (centos)
1. 使用yum包安装Apache软件 [root@Apache ~]# yum -y install httpd* [root@Apache ~]# rpm -qa | grep httpd --查 ...
- Spring Boot 如何防止重复提交?
Java技术栈 www.javastack.cn 优秀的Java技术公众号 在传统的web项目中,防止重复提交,通常做法是:后端生成一个唯一的提交令牌(uuid),并存储在服务端.页面提交请求携带这个 ...
- 创建网关项目(Spring Cloud Gateway)
创建网关项目 加入网关后微服务的架构图 创建项目 POM文件 <properties> <java.version>1.8</java.version> <s ...
- RECT,AngularJS学习网址
RECT 1.http://www.cnblogs.com/y unfeifei/p/4486125.html 2.http://www.ruanyifeng.com/blog/2015/03/rea ...
- [fw]GDT是在分段中為了相容real mode 跟 protected mode的產物
在Protected Mode下,一个重要的必不可少的数据结构就是GDT(Global Descriptor Table). 为什么要有GDT?我们首先考虑一下在Real Mode下的编程模型: 在R ...
- HDU 3571 N-dimensional Sphere( 高斯消元+ 同余 )
N-dimensional Sphere Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Oth ...
- Java的动态代理Proxy
概念梳理: 1.什么是动态代理? 答:动态代理可以提供对另一个对象的访问,同时隐藏实际对象的具体事实.代理一般会实现它所表示的实际对象的接口.代理可以访问实际对象,但是延迟实现实际对象的部分功能,实际 ...
- go语言从例子开始之Example15.闭包
Go 支持通过 闭包来使用 匿名函数.匿名函数在你想定义一个不需要命名的内联函数时是很实用的. 闭包简单理解,函数反回值是一个函数 Example: package main import " ...
- go语言从例子开始之Example25.通道方向
当使用通道作为函数的参数时,你可以指定这个通道是不是只用来发送或者接收值.这个特性提升了程序的类型安全性. Example: package main import "fmt" / ...