Codeforces 494B. Obsessive String KMP+DP

dp[i]表示到第i个可以分的段数,dp[i]=dp[i-1]+ ( dp[j]+1 ) ( 0<=j<=near[i] ) near[i]是距离i最靠右的一个含有T串的位置

B. Obsessive String

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following problem. Given a string s how many ways are there to extract k ≥ 1 non-overlapping substrings from it such that each of them contains string t as a substring? More formally, you need to calculate the number of ways to choose two sequences a1, a2, ..., ak and b1, b2, ..., bk satisfying the following requirements:

• k ≥ 1
• 
• 
• 
•   t is a substring of string saisai + 1... sbi (string s is considered as 1-indexed).

As the number of ways can be rather large print it modulo 109 + 7.

Input

Input consists of two lines containing strings s and t (1 ≤ |s|, |t| ≤ 105). Each string consists of lowercase Latin letters.

Output

Print the answer in a single line.

Sample test(s)

input

ababaaba

output

5

input

welcometoroundtwohundredandeightytwod

output

274201

input

dddd

output

12

/* ***********************************************Author        :CKbossCreated Time  :2015年03月20日 星期五 19时30分59秒File Name     :CF494B.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;const int maxn=100100;const int mod=1e9+7;char s[maxn],t[maxn];int fail[maxn];int near[maxn];int dp[maxn],sum[maxn];void getfail(char* p,int* f){int m=strlen(p);f[0]=f[1]=0;for(int i=1;i<m;i++){int j=f[i];while(j&&p[j]!=p[i]) j=f[j];f[i+1]=(p[i]==p[j])?j+1:0;}}void kmp(char* t,char* p,int* f){int n=strlen(t),m=strlen(p);getfail(p,f);int j=0;for(int i=0;i<n;i++){while(j&&p[j]!=t[i]) j=f[j];if(p[j]==t[i]) j++;if(j==m) { near[i+1]=i-m+2; j=f[j]; }}}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);scanf("%s",s); scanf("%s",t);kmp(s,t,fail);int n=strlen(s),m=strlen(t);for(int i=1;i<=n;i++)if(near[i]==0) near[i]=near[i-1];for(int i=1;i<=n;i++){dp[i]=dp[i-1];if(near[i]){dp[i]=(dp[i]+sum[near[i]-1]+near[i])%mod;}sum[i]=(sum[i-1]+dp[i])%mod;}printf("%d\n",dp[n]);        return 0;}