Another Meaning

题目链接：

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.

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.

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”, “hehe”, “hehe*”, “**”, “hehehehe”.

Source

2016 Multi-University Training Contest 4

题意：

给出字符串A和B，单词B具有两种解释，求字符串A一共有多少种意义.

题解：

考虑A中每个单词B，需要考虑它是否解释为第二种意义；很显然要用DP.

首先用kmp处理出A中的每个B，记录每个B的起点和终点(下面的代码记录的是以i为终点的B的起点，没有则为-1).

dp[i]为处理到第i个字符时共有多少种意义：

若i是某个B的终点，则dp[i] = dp[i-1] + dp[起点-1]. (前者为当前B不解释为二义，后者为把B解释为二义).

若i不是终点，则dp[i] = dp[i-1];

值得一提的是：由于背包dp需要用到初始值dp[0]，而字符串又从0开始，导致初始值为dp[-1]，所以下面的代码特别处理了这个问题.

实际上，应该在读入字符串时空出str[0]来避免这个问题.

代码：

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