Problem Description
As is known to all, in many cases, a word has two meanings. Such as “hehe”, which not only means “hehe”, but also means “excuse me”.
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.
Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.
Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.
Limits
T <= 30
|A| <= 100000
|B| <= |A|
Output
For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 1000000007.
Sample Input
4
hehehe
hehe
woquxizaolehehe
woquxizaole
hehehehe
hehe
owoadiuhzgneninougur
iehiehieh
Sample Output
Case #1: 3
Case #2: 2
Case #3: 5
Case #4: 1
Hint
In the first case, “ hehehe” can have 3 meaings: “*he”, “he*”, “hehehe”.
In the third case, “hehehehe” can have 5 meaings: “*hehe”, “he*he”, “hehe*”, “**”, “hehehehe”.
思路：递推（DP） 当前位置以前的语句的含义种数为包含当前多语义与不包含；
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const long long mod=1e9+; char s[];
char ss[];
int next1[];
long long num[];
int pos[]; void makenext1(const char P[])
{
int q,k;
int m = strlen(P);
next1[]=;
for (q = ,k = ; q < m; ++q)
{
while(k > && P[q] != P[k])
k = next1[k-];
if (P[q] == P[k])
{
k++;
}
next1[q] = k;
}
} long long calc(char T[],char P[])
{
int n,m;
int i,q;
int tot=;
n = strlen(T);
m = strlen(P);
makenext1(P);
for(i=,q = ; i < n; ++i)
{
while(q>&&P[q]!=T[i])
q=next1[q-];
if(P[q]==T[i])
{
q++;
}
if(q==m)
{
long long flag=;
pos[tot]=i-m+;
if(tot>)
{
if(pos[tot-]+m<=pos[tot])
{
num[tot]=(*num[tot-])%mod;
}
else
{
num[tot]=num[tot-]%mod;
for(int h=tot-;h>=;h--)
{
if(pos[h]+m<=pos[tot])
{
num[tot]=(num[tot]+num[h])%mod;
flag=; ///当之前不存在不相重叠的语句时;
break;
}
}
num[tot]=(num[tot]+flag)%mod;
}
}
else
{
num[tot]=;
}
tot++;
}
}
if(tot==) return ;
return num[tot-];
} int main()
{
int T;
int Case=;
cin>>T;
while(T--)
{
scanf("%s%s",s,ss);
printf("Case #%d: %lld\n",Case++,calc(s,ss));
// cout<<(calc(s,ss)%mod+mod)%mod<<endl;
}
return ;
}