開開心心學演算法--Binary String Matching
阿新 • • 發佈:2018-12-24
時間限制:3000 ms | 記憶體限制:65535 KB
難度:3
- 描述
-
Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the
pattern A appeared at the posit
- 輸入
- The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.
- 輸出
- For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
- 樣例輸入
-
3 11 1001110110 101 110010010010001 1010 110100010101011
- 樣例輸出
-
3 0 3
- 分析
- 這道題主要考察字串的配對,只要能正確分析題目意思,按照要求進行輸入輸出,就沒什麼大問題。
- 我的解:
#include <stdio.h> #include <string.h> int main() { int n,count; char a[10],b[1000],*p; scanf("%d",&n); while(n--) { count=0; scanf("%s%s",a,b); if(strlen(a)>strlen(b) || strlen(a)>10 ||strlen(b)>1000) return 0; p=b; while( (p=strstr(p,a)) !=NULL ) { count++; p++; } printf("%d\n",count); } return 1; }
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