Codeforces Round #342 (Div. 2) B. War of the Corporations 贪心
B. War of the Corporations
题目连接:
http://www.codeforces.com/contest/625/problem/B
Description
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Sample Input
intellect
tell
Sample Output
1
Hint
题意
给你s1 s2,你每次操作可以使得s1串中某个字符变成#,然后问你最小修改多少次,就可以使得s1中不含有s2子串
题解:
贪心,我们每次修改最后一个字符就好了
这样相交的一定都被修改了
代码
#include<bits/stdc++.h>
using namespace std;
string s1,s2;
int main()
{
cin>>s1>>s2;
int ans = 0;
for(int i=0;i<s1.size();i++)
{
int flag = 0;
for(int j=0;j<s2.size();j++)
{
if(i+j>s1.size())
{
flag = 1;
break;
}
if(s1[i+j]!=s2[j])
{
flag = 1;
break;
}
}
if(flag==0)
{
s1[i+s2.size()-1]='#';
ans++;
}
}
cout<<ans<<endl;
}
Codeforces Round #342 (Div. 2) B. War of the Corporations 贪心的更多相关文章
- Codeforces Round #342 (Div. 2) B. War of the Corporations(贪心)
传送门 Description A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Go ...
- Codeforces Round #342 (Div. 2)-B. War of the Corporations
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- Codeforces Round #342 (Div. 2)
贪心 A - Guest From the Past 先买塑料和先买玻璃两者取最大值 #include <bits/stdc++.h> typedef long long ll; int ...
- Codeforces Round #342 (Div. 2) B
B. War of the Corporations time limit per test 1 second memory limit per test 256 megabytes input st ...
- Codeforces Round #342 (Div. 2) D. Finals in arithmetic 贪心
D. Finals in arithmetic 题目连接: http://www.codeforces.com/contest/625/problem/D Description Vitya is s ...
- Codeforces Round #342 (Div. 2) C. K-special Tables 构造
C. K-special Tables 题目连接: http://www.codeforces.com/contest/625/problem/C Description People do many ...
- Codeforces Round #342 (Div. 2) A - Guest From the Past 数学
A. Guest From the Past 题目连接: http://www.codeforces.com/contest/625/problem/A Description Kolya Geras ...
- Codeforces Round #342 (Div. 2) E. Frog Fights set 模拟
E. Frog Fights 题目连接: http://www.codeforces.com/contest/625/problem/E Description stap Bender recentl ...
- Codeforces Round #342 (Div. 2) D. Finals in arithmetic(想法题/构造题)
传送门 Description Vitya is studying in the third grade. During the last math lesson all the pupils wro ...
随机推荐
- Delphi 提示在Delphi的IDE中,按Ctrl+Shift+G键可以为一个接口生成一个新的GUID。
对于Object Pascal语言来说,最近一段时间最有意义的改进就是从Delphi3开始支持接口(interface),接口定义了能够与一个对象进行交互操作的一组过程和函数.对一个接口进行定义包含两 ...
- 【LeetCode 239】Sliding Window Maximum
Given an array nums, there is a sliding window of size k which is moving from the very left of the a ...
- Win7+xp命令行 一键修改IP、DNS
这里提供了一个简便方法:(该方法为Win7下的,XP下的见最后一行) 第一步:新建一个txt文件 第二步:在文件中添加如下内容: netsh interface ip set address name ...
- 查找指定目录下的文件 .xml
pre{ line-height:1; color:#9f1d66; background-color:#cfe4e4; font-size:16px;}.sysFunc{color:#5d57ff; ...
- Fedora20 编译安装qemu-system
安装简介: 1.1. 本次编译安装所有的操作都在Fedora 20 x86-64上,内核版本为: 3.14.4-200.fc20.x86_64.如果在其他系统编译安装,请看其他文章. 2.安装准备: ...
- WebGIS的大众化服务
WebGIS是Web技术与GIS相结合的产物.WebGIS的真正意义在于,它将GIS从专业应用推向了大众化服务,同时为地理信息共享提供了方便而有效的途径. 一.ESRI大会关于WebGIS的报告 纵观 ...
- rm 注意
软连接ln -s lnfile file rm -rf lnfile只是删除lnfile ln -s lndir dir rm -rf lndir 删除链接 rm -rf lndir/删除目录下文件
- 搭建sql注入实验环境(基于windows)
搭建服务器环境 1.下载xampp包 地址:http://www.apachefriends.org/zh_cn/xampp.html 很多人觉得安装服务器是件不容易的事,特别是要想添加MySql, ...
- python添加windows域验证
1.安装python-ldap https://pypi.python.org/pypi/python-ldap/ 在 Ubuntu/Debian 下安装 python-ldap 模块: $ sudo ...
- create_project.py报错问题,建议用回python2.7
d:\DevTool\cocos2d-x-2.2.2\cocos2d-x-2.2.2\tools\project-creator\create_project.py 报错 d:\DevTool\coc ...