HDU 6153 A Secret

Posted 2020-10-03 十年换你一句好久不见

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

2017中国大学生程序设计竞赛 - 网络选拔赛

题目容易理解，匹配个数

#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
const int maxn=1e6+7;
ll t,la,lb,i,j,k;
char sa[maxn],sb[maxn];
ll nex[maxn],val[maxn];
void getnext()
{
    memset(nex,0,sizeof(nex));
    memset(val,0,sizeof(val));
    nex[0]=-1;
    for(i=0;i<lb;i++)
    {
        j=nex[i];
        val[i+1]=i+1;
        while(j>-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<la;i++)
    {
        j=k;
        for(k=0;j>=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     3
val    1     2     4     6     5     6     8     10    13
valfj  1     2     3+1   4+2   5     6     7+1   8+2   9+4=9+3+1::next[9]=next[3]=next[1];
*/