【47.92%】【hdu 5763】Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1440    Accepted Submission(s): 690

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.

 

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”, “he*he”, “hehe*”, “**”, “hehehehe”.
 

 

Author

FZU

 

Source

2016 Multi-University Training Contest 4

 

【题解】

设c[i]表示前i个字符能组成多少种不同意思。

c[0..lens]初值为1；

c[i] += c[i-1]-1 (不取其特殊意思)；

c[i] += c[i-lenb] (如果存在。则取其特殊意思);

abcdefgh

efgh

对于这样的输入

c[4]代表的是不取特殊意思的。==1

然后从c[4]可以推到c[4+lenb]

即c[4+lenb]+=c[4];

可以理解为

abcd****

abcdefgh这两种。

至于怎么取一个合适的位置做这样的推导。KMP

【代码】

#include <cstdio>
#include <cstring>

const int MAX_SIZE = 101000;
const int MOD = 1000000007;

int t;
int f[MAX_SIZE],c[MAX_SIZE];
char  s[MAX_SIZE], p[MAX_SIZE];
bool can[MAX_SIZE];
int lens, lenp;

void init()
{
	memset(f, 0, sizeof(f));
	memset(c, 0, sizeof(c));
	memset(can, false, sizeof(can));
}

void input_data()
{
	scanf("%s", s);
	scanf("%s", p);
	f[0] = 0; f[1] = 0;
	lens = strlen(s);lenp = strlen(p);
	for (int i = 1; i <= lenp - 1; i++)
	{
		int j = f[i];
		while (j && (p[i] != p[j])) j = f[j];
		f[i + 1] = (p[j] == p[i] ? j + 1 : 0);
	}
	int j = 0;
	for (int i = 0; i <= lens - 1; i++)
	{
		while (j && (p[j] != s[i]))j = f[j];
		if (p[j] == s[i]) j++;
		if (j == lenp)
			can[i - lenp + 1] = true;
	}
}

void get_ans()
{
	for (int i = 0; i <= lens; i++)
		c[i] = 1;
	for (int i = 0; i <= lens; i++)
	{
		if (i)
			c[i] = (c[i] + c[i - 1] - 1) % MOD;
		if (can[i])
			c[i + lenp] = (c[i+lenp] + c[i]) % MOD;
	}
}

void output_ans()
{
	printf("%d
", c[lens]);
}

int main()
{
	//freopen("F:\rush.txt", "r", stdin);
	scanf("%d", &t);
	for (int i = 1; i <= t; i++)
	{
		printf("Case #%d: ", i);
		init();
		input_data();
		get_ans();
		output_ans();
	}
	return 0;
}