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HDU - 5763 Another Meaning
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.InputThe 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|
OutputFor 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”.
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<algorithm> 5 using namespace std; 6 7 char s1[100005],s2[100005],nex[100005]; 8 long long vis[100005],dp[100005]; 9 10 int main(){ 11 long long t,i,j,cas=1; 12 scanf("%lld",&t); 13 while(t--){ 14 scanf("%s%s",s1,s2);// s2 是匹配串 15 memset(nex,-1,sizeof(nex)); //初始化为-1 16 memset(vis,0,sizeof(vis)); 17 memset(dp,0,sizeof(dp)); 18 long long n=strlen(s1); 19 long long m=strlen(s2); 20 j=-1; 21 for(i=1;i<m;i++){ 22 while(j>=0&&s2[j+1]!=s2[i])j=nex[j]; 23 if(s2[j+1]==s2[i])j++; 24 nex[i]=j; 25 } 26 j=-1; 27 for(i=-1;i<n-1;i++){ 28 while(j>=0&&s2[j+1]!=s1[i+1])j=nex[j]; 29 if(s2[j+1]==s1[i+1])j++; 30 if(j==m-1){ 31 //j=next[j]; //重叠匹配 32 //j=-1; //从头匹配 33 j=nex[j]; 34 vis[i+1]=1; 35 } 36 } 37 long long num=0; 38 for(i=0;i<m;i++){ 39 dp[i]=vis[i]; 40 } 41 for(;i<n;i++){ 42 if(vis[i]){ 43 dp[i]=(dp[i-1]+dp[i-m]+1)%1000000007; 44 } 45 else dp[i]=dp[i-1]; 46 } 47 printf("Case #%lld: %lld\n",cas++,(dp[n-1]+1)%1000000007); 48 } 49 return 0; 50 }
HDU - 5763 Another Meaning