Educational Codeforces Round 8 B 找规律
1 second
256 megabytes
standard input
standard output
Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.
You are given a string s consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.
A substring of a string is a nonempty sequence of consecutive characters.
For example if string s is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
The only line contains string s (1 ≤ |s| ≤ 3·105). The string s contains only digits from 0 to 9.
Print integer a — the number of substrings of the string s that are divisible by 4.
Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
124
4
04
3
5810438174
9 题意:给你一个串 询问子串中可以除尽4的个数 子串允许有前导‘0’ 题解:1.首先对于单个字符 ‘0’ ‘4’ ‘8’可以被除进
2.对于两个连续字符 组成的数exm=(s[i]-'0')+(s[i-1]-'0')*10必须除尽4
3.我们知道100能够整除4 所以只需要统计 以 (满足条件的两个连续字符)为后缀的子串的个数
求和
#include<bits/stdc++.h>
#define ll __int64
#define mod 1e9+7
#define PI acos(-1.0)
#define bug(x) printf("%%%%%%%%%%%%%",x);
#define inf 1e8
using namespace std;
#define ll long long
char s[];
map<int,int> mp;
ll ans;
int main()
{
scanf("%s",s);
ans=;
for(int i=;i<=;i+=)
mp[i]=;
int len=strlen(s);
for(int i=;i<len;i++)
{
if((s[i]-'')%==)
ans++;
int exm=(s[i]-'')+(s[i-]-'')*;
if(mp[exm])
{
ans=ans+i;
}
}
printf("%I64d\n",ans);
return ;
}
Educational Codeforces Round 8 B 找规律的更多相关文章
- Educational Codeforces Round 85 (Rated for Div. 2)
\(Educational\ Codeforces\ Round\ 85\ (Rated\ for\ Div.2)\) \(A. Level Statistics\) 每天都可能会有人玩游戏,同时一部 ...
- Educational Codeforces Round 37
Educational Codeforces Round 37 这场有点炸,题目比较水,但只做了3题QAQ.还是实力不够啊! 写下题解算了--(写的比较粗糙,细节或者bug可以私聊2333) A. W ...
- Educational Codeforces Round 5
616A - Comparing Two Long Integers 20171121 直接暴力莽就好了...没什么好说的 #include<stdlib.h> #include&l ...
- Educational Codeforces Round 35 (Rated for Div. 2)
Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...
- Educational Codeforces Round 48 (Rated for Div. 2) CD题解
Educational Codeforces Round 48 (Rated for Div. 2) C. Vasya And The Mushrooms 题目链接:https://codeforce ...
- Educational Codeforces Round 59 (Rated for Div. 2) DE题解
Educational Codeforces Round 59 (Rated for Div. 2) D. Compression 题目链接:https://codeforces.com/contes ...
- Educational Codeforces Round 58 (Rated for Div. 2) 题解
Educational Codeforces Round 58 (Rated for Div. 2) 题目总链接:https://codeforces.com/contest/1101 A. Min ...
- Educational Codeforces Round 32
http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 ...
- Educational Codeforces Round 35 A. Nearest Minimums【预处理】
[题目链接]: Educational Codeforces Round 35 (Rated for Div. 2) A. Nearest Minimums time limit per test 2 ...
随机推荐
- 课堂使用的Linux命令
1.ls 显示目录文件 2.ls -l = ll 显示 3.vi 创建于编辑 4.mv 改名 5. ./ 当前目录 6.chmod 修改权限 4 读 2 编辑 1 运行 7.去除UI界面 syste ...
- 更新MySQL数据库( java.sql.SQLException: No value specified for parameter 1) 异常 解决方法
package com.swift; import java.io.File; import java.sql.Connection; import java.sql.PreparedStatemen ...
- Linux下C程序内存泄露检测
在linux下些C语言程序,最大的问题就是没有一个好的编程IDE,当然想kdevelop等工具都相当的强大,但我还是习惯使用kdevelop工具,由于没有一个习惯的编程IDE,内存检测也就成了在lin ...
- 博学谷-数据分析numpy
import numpy as np print np.version.version np.array([1,2,3,4]) np.arange(15) np.array(range(10)) = ...
- pycahrm git配置笔记
1. 在file - setting - plugins 中查看是否有github插件, 此处是用于处理插件位置
- 三十二、MySQL 导出数据
MySQL 导出数据 MySQL中你可以使用SELECT...INTO OUTFILE语句来简单的导出数据到文本文件上. 使用 SELECT ... INTO OUTFILE 语句导出数据 以下实例中 ...
- nginx安装php环境
1.php下载地址 https://secure.php.net/downloads.php(此次安装版本为7.0.33) 2.安装依赖的包 yum -y install libxml2 yum -y ...
- PKI
公钥基础设施(Public Key Infrastructure,简称PKI)是眼下网络安全建设的基础与核心,是电子商务安全实施的基本保障,因此,对PKI技术的研究和开发成为眼下信息安全领域的热点.本 ...
- 扒一扒 EventServiceProvider 源代码
Ajax用一句话来说就是无须刷新页面即可从服务器取得数据.注意,虽然Ajax翻译过来叫异步JavaScript与XML,但是获得的数据不一定是XML数据,现在服务器端返回的都是JSON格式的文件. 完 ...
- 水题:HDU-1088-Write a simple HTML Browser(模拟题)
解题心得: 1.仔细读题,细心细心...... 2.题的几个要求:超过八十个字符换一行,<br>换行,<hr>打印一个分割线,最后打印一个新的空行.主要是输出要求比较多. 3. ...