A Secret

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 2523    Accepted Submission(s): 934

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.

题目链接：HDU 6153

扩展KMP解决的是对于每一个每一个$S$的后缀$Suffix(i)$，它与$T$串最长的公共前缀是多少，即它与$T$串各自从头开始能匹配几个，一旦有一个不匹配即结束。

这题求的是每一个$S_2$的后缀出现在$S_1$有几次，那么我们可以把两个串都反过来变成求$S_2$的前缀在$S_1$出现过几次，这样就可以发现如果$S_2$的前缀$prefix(j)$与$S_1$的后缀$Suffix(i)$的最长公共前缀为$k$，那么包括$prefix(j)$以及$j$以前的前缀均在$S_1$的$i$处出现过一次，贡献为$(k+1)*k/2$，那么这样求一遍$extend[]$数组后统计答案即可

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1e6 + 7;
const LL mod = 1000000007LL;
char s[N], t[N];
int nxt[N], ex[N];

void getnxt(char T[], int len, int nxt[])
{
    nxt[0] = len;
    int a;
    int p;
    for (int i = 1, j = -1; i < len; i++, j--)
    {
        if (j < 0 || i + nxt[i - a] >= p)
        {
            if (j < 0)
                p = i, j = 0;
            while (p < len && T[p] == T[j])
                p++, j++;
            nxt[i] = j;
            a = i;
        }
        else
            nxt[i] = nxt[i - a];
    }
}
void getextend(char S[], char T[], int ls, int lt, int extend[], int nxt[])
{
    getnxt(T, lt, nxt);  //得到next
    int a;
    int p;             //记录匹配成功的字符的最远位置p，及起始位置a
    for (int i = 0, j = -1; i < ls; i++, j--)  //j即等于p与i的距离，其作用是判断i是否大于p（如果j<0，则i大于p）
    {
        if (j < 0 || i + nxt[i - a] >= p)  //i大于p（其实j最小只可以到-1，j<0的写法方便读者理解程序），
        {   //或者可以继续比较（之所以使用大于等于而不用等于也是为了方便读者理解程序）
            if (j < 0)
                p = i, j = 0;  //如果i大于p
            while (p < ls && j < lt && S[p] == T[j])
                p++, j++;
            extend[i] = j;
            a = i;
        }
        else
            extend[i] = nxt[i - a];
    }
}
int main(void)
{
    int T, i;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%s%s", s, t);
        int la = strlen(s);
        int lb = strlen(t);
        reverse(s, s + la);
        reverse(t, t + lb);
        getnxt(t, lb, nxt);
        getextend(s, t, la, lb, ex, nxt);
        LL ans = 0;
        for (i = 0; i < la; ++i)
        {
            LL temp = (LL)ex[i] * ((LL)ex[i] + 1LL) / 2LL;
            ans = (ans + temp) % mod;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}