acc suffix print amp blank 拓展 nbsp rod div

A Secret

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 975 Accepted Submission(s): 372

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.

題意

給你兩個字符串A,B，現在要你求B串的後綴在A串中出現的次數和後綴長度的乘積和為多少。

思路

擴展KMP模板題，將s和t串都逆序以後就變成了求前綴的問題了，擴展KMP求處從i位置開始的最長公共前綴存於數組，最後通過將數組的值不為0的進行一個等差數列和的和就可以了。

代碼：

 1 #include <iostream>
 2 #include <string>
 3 #include <string.h>
 4 #include <cstring>
 5 #include <algorithm>
 6 using namespace std;
 7 const int maxn = 1e6 + 10;
 8 const int mod = 1e9 + 7;
 9 typedef long long ll;
10 int cnt[maxn];
11 char A[maxn],B[maxn];
12 int Next[maxn],ex[maxn];
13 ll add(ll n)
14 {
15     ll m=((n%mod)*((n+1)%mod)/2)%mod;
16     return m;
17 }
18 void kmp(char P[])
19 {
20     int m=strlen(P);
21     Next[0]=m;
22     int j=0,k=1;
23     while(j+1<m&&P[j]==P[j+1]) j++;
24     Next[1]=j;
25     for(int i=2; i<m; i++)
26     {
27         int p=Next[k]+k-1;
28         int L=Next[i-k];
29         if(i+L<p+1) Next[i]=L;
30         else
31         {
32             j=max(0,p-i+1);
33             while(i+j<m&&P[i+j]==P[j])
34                 j++;
35             Next[i]=j;
36             k=i;
37         }
38     }
39 }
40 
41 void exkmp(char P[],char T[])
42 {
43     int m=strlen(P),n=strlen(T);
44     kmp(P);
45     int j=0,k=0;
46     while(j<n&&j<m&&P[j]==T[j])
47         j++;
48     ex[0]=j;
49     for(int i=1; i<n; i++)
50     {
51         int p=ex[k]+k-1;
52         int L=Next[i-k];
53         if(i+L<p+1)
54             ex[i]=L;
55         else
56         {
57             j=max(0,p-i+1);
58             while(i+j<n&&j<m&&T[i+j]==P[j])
59                 j++;
60             ex[i]=j;
61             k=i;
62         }
63     }
64 }
65 
66 int main()
67 {
68     int t;
69     scanf("%d",&t);
70     while(t--)
71     {
72         scanf("%s%s",A,B);
73         int lenA=strlen(A);
74         int lenB=strlen(B);
75         reverse(A,A+lenA);
76         reverse(B,B+lenB);
77         kmp(B);
78         memset(Next,0,sizeof(Next));
79         memset(ex,0,sizeof(ex));
80         exkmp(B,A);
81         ll ans = 0;
82         for(int i=0;i<lenA;i++)
83         {
84             if(ex[i])
85                 ans=(ans+add(ex[i])%mod)%mod;
86         }
87         printf("%lld\n",ans%mod);
88     }
89     return 0;
90 }

HDU 6153 拓展KMP （2017CCPC）