Hdu 5763 another meaning hash +dp

another meaning

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 917 Accepted Submission (s): 434

problem Descriptionas is known to all, in many cases, a word has both 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??.?? are so smart and he knows the word B in the Sentence has a meanings. He wants to know how many kinds of meanings meizi can express. 

InputThe first line of the input gives the number of test cases T; T test cases follow.
Each test case contains-strings A and B, a means the sentence Meizi sends to??, B means the word B-which has a Menaings. String only contains lowercase letters.

Limits
T <=
| a| <= 100000
| b| <= | a|

 

Outputfor each test case, output one line containing ' case #x: Y ' (without quotes), where x is the ' test Case number (Starti Ng from 1) and Y are the number of the different meaning of this sentence. Since This number is quite large, you should output the answer modulo 1000000007. 

Sample Input4hehehehehewoquxizaolehehewoquxizaoleheheheheheheowoadiuhzgneninouguriehiehieh 

Sample OutputCase #1:3 case #2:2 case #3:5 case #4:1 HintIn the first case , "hehehe" Can has 3 meaings: "*he", "he*", "hehehe". In the third case, "Hehehehe" Can has 5 meaings: "*hehe", "He*he", "hehe*", "* *", "hehehehe".

Test instructions: give you a substring of the main string, and then match the string to the substring to change the match to * (or not to change). Ask the main string how many clocks are different;

#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <vector > #include <queue> #include <cstring> #include <string> #include <algorithm>using namespace Std;typedef Long long ll;typedef unsigned long long ull; #define MM (A, B) memset (A,b,sizeof (a)); #define INF 0x7f7f7f7f#def ine for (i,n) for (int i=1;i<=n;i++) #define CT continue; #define PF printf#define SC scanfconst int Mod=1000000007;const i   NT N=1e5+10;int la,lb;ll dp[n];char a[n],b[n];ull ha[n],hb;ull seed=13331;void init () {hb=0;   for (int i=0;i<lb;i++) hb=hb*seed+b[i];   Ull base=1;   for (int i=1;i<=lb-1;i++) base*=seed;   HA[0]=A[0];   for (int i=1;i<=lb-1;i++) ha[i]=ha[i-1]*seed+a[i]; for (int i=lb;i<la;i++) ha[i]= (ha[i-1]-a[i-1-(lb-1)]*base) *seed+a[i];}    int main () {int cas,kk=0;    scanf ("%d", &cas);        while (cas--) {scanf ("%s", a);        scanf ("%s", b);        La=strlen (a); Lb=strlen (b); if (la<lb) {PRintf ("Case #%d:1\n", ++KK);        CT;}        Init ();        for (int i=0;i<=lb-1;i++) dp[i]=1;        if (HA[LB-1]==HB) dp[lb-1]=2;           for (int i=lb;i<la;i++) {dp[i]=dp[i-1]%mod;        if (HA[I]==HB) dp[i]= (dp[i]+dp[i-lb])%mod;    } printf ("Case #%d:%lld\n", ++kk,dp[la-1]%mod); } return 0;}

Analysis: Error point:

1.BKDRhash is not very skilled, only the original final output of the form of &, resulting in the final computational complexity;

Improved: Bkdrhash form: Pre*seed+a[i],seed is a large prime number such as 13331, pre is the hash value before I;

A[i] is the character

2, do not think of DP, current subscript i, dp[i] value, if the current I do not match to the substring, dp[i]=dp[i-1];

If matched to a substring, dp[i]=dp[i-1]+dp[i-lb];

Hdu 5763 another meaning hash +dp