Binary String Matching

问题 B: Binary String Matching

时间限制: 3 Sec  内存限制: 128 MB
提交: 4  解决: 2
[提交][状态][讨论版]

题目描述

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

输入

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.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main()
{
    int n;
    char pattern[1111];
    char str[1111];
    char temp[1111];
    int cou=0;
    int b;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        cou=0;
        scanf("%s",pattern);
        scanf("%s",str);
        int lens=strlen(str);
        int lenp=strlen(pattern);
        for(int j=0;j<lens;j++){
            b=0;
            if(str[j]==pattern[0]){
                for(int h=0;h<lenp;h++){
                    if(pattern[h]!=str[j+h]){
                        b=1;
                        break;
                    }else{
                        continue;
                    }
                }
                if(b==0){
                    cou++;
                }
            }
        }
        if(i==n-1){
            printf("%d",cou);
        }else{
            printf("%d
",cou);
        }




    }

    return 0;
}

 　

只前在自己学校oj提交可以ac，但到别的oj提交就过不了，自己学校oj的测试数据太弱！！

又重新写了一个

#include <iostream>
#include <cstdio>
#include <stack>
#include <cstring>

using namespace std;

int main()
{
    int n;
    char ch;
    int b=0;
    char s[10005];
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        stack<char> st1;
        scanf("%s",s);
        int len=strlen(s);
        st1.push(s[0]);
        for(int j=1;j<len;j++){
            b=0;
            if(st1.empty()){
                st1.push(s[j]);
            }else{
                if(s[j]==')'){
                    if(st1.top()=='('){
                        st1.pop();
                    }else{
                        printf("No
");
                        b=1;
                        while(!st1.empty()){
                            st1.pop();
                        }
                        break;
                    }
                }
                if(s[j]==']'){
                    if(st1.top()=='['){
                        st1.pop();
                    }else{
                        printf("No
");
                        b=1;
                        while(!st1.empty()){
                            st1.pop();
                        }
                        break;
                    }
                }
                if(s[j]=='['||s[j]=='('){
                    st1.push(s[j]);
                }
            }

        }

        if(b!=1&&st1.empty()){
            printf("Yes
");
        }
        if(b==0&&!st1.empty()){
            printf("No
");
        }
    }
    return 0;
}