这道题的官方题解是dp，但是可以暴力出来。改天再研究怎么dp。

暴力的时候，如果计算sum的时候，调用strlen函数会超时，可见这个函数并不是十分的好。以后能不用尽量不用。

La Vie en rose

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 861    Accepted Submission(s): 461

Problem Description

Professor Zhang would like to solve the multiple pattern matching problem, but he only has only one pattern string p=p1p2...pm. So, he wants to generate as many as possible pattern strings from p using the following method:

1. select some indices i1,i2,...,ik such that 1≤i1<i2<...<ik<|p| and |ij−ij+1|>1 for all 1≤j<k.
2. swap pij and pij+1 for all 1≤j≤k.

Now, for a given a string s=s1s2...sn, Professor Zhang wants to find all occurrences of all the generated patterns in s.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers n and m (1≤n≤105,1≤m≤min{5000,n}) -- the length of s and p.

The second line contains the string s and the third line contains the string p. Both the strings consist of only lowercase English letters.

Output

For each test case, output a binary string of length n. The i-th character is "1" if and only if the substring sisi+1...si+m−1 is one of the generated patterns.

Sample Input

3
4 1
abac
a
4 2
aaaa
aa
9 3
abcbacacb
abc

Sample Output

1010
1110
100100100

Author

zimpha

Source

2016 Multi-University Training Contest 2

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstring>
#include<string>
using namespace std;
const int maxx=100005;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m;
        char s[maxx];
        char p[maxx];
        scanf("%d%d",&n,&m);
        scanf("%s",s);
        scanf("%s",p);
        int sump[maxx];
        int sums[maxx];
        sump[0]=sums[0]=0;
        for(int i=1;i<=n;i++){
            sums[i]=s[i-1]-‘a‘+sums[i-1];
        }
        for(int i=1;i<=m;i++){
            sump[i]=p[i-1]-‘a‘+sump[i-1];
        }
        int pis=sump[m];
        int shave=sums[m];
        for(int i=m;i<=n;i++){
            shave=sums[i]-sums[i-m];
           // cout<<"shave: "<<shave<<endl;
            if(pis==shave){
                int sta=i-m;
                int pos=sta;
                int flag=1;
                for(int j=0;pos<i;j++){
                    if(s[pos]==p[j]){
                        pos++;
                    }else{
                        if(pos+1<i&&s[pos+1]==p[j]&&s[pos]==p[j+1]){
                            pos+=2;
                            j++;
                        }else{
                            flag=0;
                            break;
                        }
                    }
                }
                if(flag){
                    printf("1");
                }else{
                    printf("0");
                }
            }else{
                printf("0");
            }
        }
        for(int i=m-1;i>0;i--){
            printf("0");
        }
        printf("\n");

    }
    return 0;
}