nyoj 5 Binary String Matching
阿新 • • 發佈:2018-04-24
for each 題目 pri span std example clr ask spl
Binary String Matching
時間限制:3000 ms | 內存限制: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
/** 題目大意: 求一個串的子串,再判斷該串與題目所給的串是否相同 分析: 通過C++提供的substr函數求子串 substr用法:str2 = str1.substr (index, length) **/
C/C++代碼實現:
#include <iostream> #include <algorithm> #include <cmath> #include
nyoj 5 Binary String Matching