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. Android--ListView与数据绑定(Xamarin)

    ListView 控件是一个条目容器, 用于显示集合对象(如数组, List<T>, ObservableCollection<T>等)的每一个条目, 并提供滚动功能. 列表视 ...

  2. Sourceinsight最佳配色方案及颜色字体调整方法

    在Ubuntu下面用Gedit有一款比较好看的配色,应该是Darkblue.按照那个样子在SI里面做了一个差不多的,按个人喜好,背景色换成黑色,如下所示: 配色的方案文件可以从此处链接免费下载: 配色 ...

  3. iOS常见算法(二分法 冒泡 选择 快排)

    二分法: 平均时间复杂度:O(log2n) int halfFuntion(int a[], int length, int number)  { int start = 0; int end = l ...

  4. Linux TOP命令 按内存占用排序和按CPU占用排序

    P – 以 CPU 占用率大小的顺序排列进程列表M – 以内存占用率大小的顺序排列进程列表 http://blog.csdn.net/xiliuhu/article/details/6449377

  5. [SharePoint 2010] Copy list item with version history and attachment

    private void MoveItem(SPListItem sourceItem, SPListItem destinationItem) { if (sourceItem == null || ...

  6. @overrive报错解决方案

    有时将项目移到另一个环境里出现@overrive报错的情况,可以看看eclipse工具的编译设置,将编译版本设置高点.

  7. STMFD 和LDMFD指令

    http://blog.163.com/oy_mcu/blog/static/16864297220120193458892/ LDM/STM指令主要用于现场保护,数据复制,参数传送等. STMFD指 ...

  8. C#模拟Http与Https请求框架实例

    using System.Text; using System.Net; using System.IO; using System.Text.RegularExpressions; using Sy ...

  9. Computop支付网关(一) credit Card

    1.界面没有中文,只能选择英文 sp – Spanish; en – English; ca – Catalan; fr – French; de – German; du – Dutch; it – ...

  10. 用Canvas制作简单的画图工具

    今天用Canvas制作了一个画图工具,非常简单,功能也不是很多,主要有背景网格,画线,画圆,画矩形和画圆角矩形,也用到了canvas的一些基本知识,在这里一一列举. 1.线段的绘制: 如何绘制真正的1 ...