字符串处理Keep Deleting

最新推荐文章于 2020-08-28 13:40:34 发布

原创最新推荐文章于 2020-08-28 13:40:34 发布 · 526 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#字符串处理

字符串处理专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个基于字符串匹配的算法，该算法通过不断从目标字符串B中删除子字符串A，直至A不再是B的子串，最终输出删除次数。采用数组模拟栈的方式实现，适用于处理大量字符串匹配与删除操作。

Assume that string A is the substring of string B if and only if we can find A in B, now we have a string A and another string B, your task is to find a A in B from B's left side to B's right side, and delete it from B till A is not a substring of B, then output the number of times you do the delete.

There are only letters(A-Z, a-z) in string A and B

This problem contains multiple test cases. Each case contains two line, the first line is string A, the second line is string B, the length of A is less than 256, the length of B is less than 512000

Print exactly one line with the number of times you do the delete for each test case.

Sample Input

abcd
abcabcddabcdababcdcd

Sample Output

Hint

abcabcddabcdababcdcd delete=0
abcdabcdababcdcd     delete=1
abcdababcdcd         delete=2
ababcdcd             delete=3
abcd                 delete=4
                     delete=5

意思是给你两个字符串A， B删掉B内存在的A串，删掉后就会连接起来了，看最后要删几次。

捉狂了，就是要用数组模拟栈，如果长度够是了lena的话看是否能删不能继续但是下一是所在字符串的后lena这段时候合适，能就删，

#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
const int base =512010;
char s[base],A[base],B[base],c[base];
int main()
{
   while(~scanf("%s%s", A, B))
   {
    int la=strlen(A);
    int lb=strlen(B);
    memset(s, 0, sizeof(s));
    int cout=0,k=0;
    for(int i =0; i<lb; i++)
    {
       s[k++]=B[i];
       if(k>=la)
       {
           s[k]='\0';
           if(strcmp(s+k-la,A)==0) //就是看在s串里面最后的la个这个子串是否和A相同
           {
               k-=la;
               cout++;
           }
       }
    }
    printf("%d\n", cout);
   }
    return 0;
}