【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\n", 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;
}