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.
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|
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: 1HintIn 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
【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;
}