time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output


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 test(s)

input

intellect

tell

output

1

input

google

apple

output

0

input

sirisiri

sir

output

2

Note

In the first sample AI’s name may be replaced with “int#llect”.

In the second sample Gogol can just keep things as they are.

In the third sample one of the new possible names of AI may be “s#ris#ri”.

在原字符中至少增添多少个#使得没有第二个字符串,暴力即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <map>
#include <list>
#include <vector>
#include <functional>
#include <algorithm>
using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int>PII; typedef vector<int>VI; typedef vector<LL>VL; const int INF = 0x3f3f3f3f; const double eps = 1e-6; const double Pi = acos(-1.0); char str[100100]; char c[40]; int main()
{
scanf("%s %s",str,c); int len = strlen(str); int ans = 0; for(int i=0;i<len;i++)
{
if(str[i]==c[0])
{
int k;
for( k=0;c[k]!='\0';k++)
{
if(str[i+k]!=c[k])
{
break;
}
} if(c[k]=='\0')
{
ans++; i=i+k-1;
}
}
} printf("%d\n",ans);
return 0;
}

Codeforces Round #342 (Div. 2)-B. War of the Corporations的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. 前端页面使用 Json对象与Json字符串之间的互相转换

    前言 在前端页面很多时候都会用到Json这种格式的数据,最近没有前端,后端的我也要什么都要搞,对于Json对象与Json字符串之间的转换终于摸清楚了几种方式,归纳如下! 一:Json对象转换为json ...

  2. python零碎知识点一

    dir(object),列出对象所有可以用的的方法(参数可以为任意对象,例如class,func等) >>>dir('str') ['__add__', '__class__', ' ...

  3. 使用async 和 await方法来

    先看直接的代码请求方式地啊: 这里是我们同步方法的实现: using System; using System.Collections.Generic; using System.Diagnostic ...

  4. shell中对字符串的处理

    1.替换字符串1为字符串2 sed "s/str1/str2/g" 2.获取字符串中的一部分 例:boke-blade 取得boke:sed -e "s/-.*//g&q ...

  5. vim - Convert between hex and decimal

    http://vim.wikia.com/wiki/VimTip448 ga g8

  6. python中使用正则表达式

    正则表达式元字符:. ^ $ * + ? {} [] \ | ()第一部分:1.[] 常用来指定一个字符集,用于匹配其中的一个字符:^,$元字符在里面不起作用,但是+-*等符号在[]中还是有特殊含义的 ...

  7. ActiveMQ 复杂类型的发布与订阅

    很久没po文章了,但是看到.Net里关于ActiveMQ发送复杂类型的文章确实太少了,所以贴出来和大家分享 发布: //消息发布 public class Publisher { private IC ...

  8. WebForm 内置对象2

    Session: 与Cookies相比 相同点:每一台电脑访问服务器,都会是独立的一套session,key值都一样,但是内容都是不一样的 以上所有内容,都跟cookies一样 不同点: 1.Sess ...

  9. 开发Windows Phone应用程序之后的感觉

    刚刚历时一个多月完成了酒美网(我之前的公司)Windows Phone版客户端,发现自己的自学能力还可以,但是还是有好多东西摸不清,到今天我才刚刚对MVVM入门,更对MVVMLight这个框架有进一步 ...

  10. 00Linux学习及角色定义

    一.嵌入式Linux学习顺序 二.Linux架构 三.Linux工程师角色划分 四. 嵌入式应用工程师工作内容与所需知识点 从图 1可以知道, 嵌入式应用工程师主要从事与产品相关的嵌入式 Linux ...