Binary String Matching

時(shí)間限制：3000?ms ?|? 內(nèi)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

來源

網(wǎng)絡(luò)

上傳者

naonao

?kmp ?裸題；

#include <cstdio> #include <cstring> #include <iostream> using namespace std; char a[15], b[1010]; int p[1010], lena, lenb, ans; void Getp() {int i = 0, j = -1;p[i] = j;while(i != lena){if(j == -1 || a[i] == a[j]){i++, j++;p[i] = j; } elsej = p[j];} } void Kmp() {Getp();int i = 0, j = 0;while(i != lenb){if(j == -1 || b[i] == a[j]) i++, j++;else j = p[j];if(j == lena)ans++; } printf("%d\n", ans); } int main() {int t;scanf("%d", &t);while(t--){ans = 0;scanf("%s%s", a, b);lena = strlen(a);lenb = strlen(b);Kmp();}return 0; }

?

暴力： 竟然也是 0 ms;?

#include <cstdio> #include <cstring> #include <iostream> using namespace std; char a[15], b[1010]; int lena, lenb; int main() {int t;scanf("%d", &t);while(t--){int i, j, sum = 0;scanf("%s %s", a, b);int lena = strlen(a), lenb = strlen(b);for(i = 0; i <= lenb - lena; i++){if(b[i] == a[0]){for(j = 1; j < lena; j++)if(b[i+j] != a[j])break;if(j == lena)sum++;} }printf("%d\n", sum);} return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/soTired/p/4749391.html

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