hdu 5745 La Vie en rose(2016多校第二场)
La Vie en rose
Time Limit: 14000/7000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 643 Accepted Submission(s):
328
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 .
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.
n . The i -th character is "1" if and only if the substring sisi+1...si+m−1 is one of the generated patterns.
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define ll long long
using namespace std;
char s1[],s2[];
int main()
{
int T,i,j;
scanf("%d",&T);
while(T--)
{
int len1,len2;
memset(s1,'\0',sizeof(s1));
memset(s2,'\0',sizeof(s2));
scanf("%d%d",&len1,&len2);
scanf("%s%s",s1,s2);
char w;
if(len1<len2)
{
for(i=; i<len1; i++)
printf("");
printf("\n");
continue;
}
for(i=; i<len1; i++)
{
int t=;
for(j=i; j<i+len2&&j<len1; j++)
{
if(s1[j]!=s2[t])
{
if(j==i+len2-)
break;
if(s1[j+]!=s2[t]||s1[j]!=s2[t+])
break;
else
{
t+=;
j+=;
}
}
else
{
t++;
}
}
if(j==i+len2)
printf("");
else
printf(""); }
printf("\n");
}
return ;
}
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