字符串处理Keep Deleting

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

abcdabcabcddabcdababcdcd

Sample Output

5

Hint

abcabcddabcdababcdcd delete=0abcdabcdababcdcd     delete=1abcdababcdcd         delete=2ababcdcd             delete=3abcd                 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;}