A Secret

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 256000/256000 K (Java/Others)

Total Submission(s): 817 Accepted Submission(s): 321

Problem Description

Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell:

Suffix(S2,i) = S2[i...len].Ni is the times that Suffix(S2,i) occurs in S1 and Li is the length of Suffix(S2,i).Then the secret is the sum of the product of Ni and Li.

Now SF wants you to help him find the secret.The answer may be very large, so the answer should mod 1000000007.

Input

Input contains multiple cases.

The first line contains an integer T,the number of cases.Then following T cases.

Each test case contains two lines.The first line contains a string S1.The second line contains a string S2.

1<=T<=10.1<=|S1|,|S2|<=1e6.S1 and S2 only consist of lowercase ,uppercase letter.

Output

For each test case,output a single line containing a integer,the answer of test case.

The answer may be very large, so the answer should mod 1e9+7.

Sample Input

2

      aaaaa 

aa

abababab

aba

Sample Output

13

      19 
Hint
 case 2: Suffix(S2,1) = "aba", Suffix(S2,2) = "ba", Suffix(S2,3) = "a". N1 = 3, N2 = 3, N3 = 4. L1 = 3, L2 = 2, L3 = 1. ans = （3*3+3*2+4*1）%1000000007. 

Source

题目容易理解，匹配个数

#include 
      
        #include 
       
         #include 
        
          #include 
         
          #include 
          
            #include 
           
             #include 
            
              #include 
             
              #include 
              
               #include 
               
                #include 
                
                  #include 
                  #include 
                  
                    using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x
                   
                    -1) { if(sb[j]==sb[i]) {nex[i+1]=j+1,val[i+1]+=val[j+1],val[i+1]%=MOD;break;} j=nex[j]; } } /*for(i=1;i<=lb;i++) { printf("%lld ",val[i]); } printf("\n");*/}ll matchfind(){ ll ans=0; k=0; for(i=0;i
                    
                     =0;j=nex[j]) { if(sa[i]==sb[j]) { k=j+1; break; } } ans=(ans+val[k])%MOD; } return ans;}int main(){ scanf("%lld",&t); while(t--) { scanf("%s %s",sa,sb); la=strlen(sa),lb=strlen(sb); reverse(sa,sa+la),reverse(sb,sb+lb); getnext(); printf("%lld\n",matchfind()); } return 0;}/* a b a b b b a b a next 0 0 1 2 0 0 1 2 3val 1 2 4 6 5 6 8 10 13valfj 1 2 3+1 4+2 5 6 7+1 8+2 9+4=9+3+1::next[9]=next[3]=next[1];*/

转载于:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7399163.html

你可能感兴趣的文章

[Ocean Modelling for Begineers] Ch3. Basics of Geophysical Fluid Dynamics