Classic Quotation

Problem Description

When online chatting, we can save what somebody said to form his ‘‘Classic Quotation‘‘. Little Q does this, too. What‘s more? He even changes the original words. Formally, we can assume what somebody said as a string S whose length is n. He will choose a continuous substring of S(or choose nothing), and remove it, then merge the remain parts into a complete one without changing order, marked as S′. For example, he might remove ‘‘not‘‘ from the string ‘‘I am not SB.‘‘, so that the new string S′ will be ‘‘I am SB.‘‘, which makes it funnier.

After doing lots of such things, Little Q finds out that string T occurs as a continuous substring of S′ very often.

Now given strings S and T, Little Q has k questions. Each question is, given L and R, Little Q will remove a substring so that the remain parts are S[1..i] and S[j..n], what is the expected times that T occurs as a continuous substring of S′ if he choose every possible pair of (i,j)(1≤i≤L,R≤j≤n) equiprobably? Your task is to find the answer E, and report E×L×(n−R+1) to him.

Note : When counting occurrences, T can overlap with each other.

Input

The first line of the input contains an integer C(1≤C≤15), denoting the number of test cases.

In each test case, there are 3 integers n,m,k(1≤n≤50000,1≤m≤100,1≤k≤50000) in the first line, denoting the length of S, the length of T and the number of questions.

In the next line, there is a string S consists of n lower-case English letters.

Then in the next line, there is a string T consists of m lower-case English letters.

In the following k lines, there are 2 integers L,R(1≤L<R≤n) in each line, denoting a question.

Output

For each question, print a single line containing an integer, denoting the answer.

Sample Input

1
8 5 4
iamnotsb
iamsb
4 7
3 7
3 8
2 7

Sample Output

1
1
0
0

题意：

　　给两个字符串只包含小写字母，长度分别为n,m

　　k个询问，每次询问给出一个L,R

　　任意的 ( i , j ) ( 1 ≤ i ≤ L , R ≤ j ≤ n ) 删除S串范围(i+1,j-1)内的字符，求出T串在新串内出现的次数总和

题解：

　　我还是照搬官方题解吧

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 6e4+10, M = 2e2+20,inf = 2e9;

int zfail[N],ffail[N];
LL dp[N][M],f[N][M],sumdp[N][M],sumf[N][M],dp2[N][M],f2[N][M];
char a[N],b[N];
int n,m,k,T;
LL solve(int ll,int rr) {
    LL ret = 0;
    ret += 1LL * sumdp[ll][m] * (n - rr + 1) + 1LL * sumf[rr][1] * (ll);
    for(int i = 1; i < m; ++i) {
        ret += 1LL*dp2[ll][i] * f2[rr][i+1];
    }
    return ret;
}
void init() {
    for(int j = 0; j <= m+1; ++j) zfail[j] = 0,ffail[j] = m+1;
    for(int i = 0; i <= n+1; ++i)
        for(int j = 0; j <= m+1; ++j)
            dp[i][j] = 0,f[i][j] = 0,sumdp[i][j] = 0,sumf[i][j] = 0;
    int j = 0;
    for(int i = 2; i <= m; ++i) {
        while(j&&b[j+1]!=b[i]) j = zfail[j];
        if(b[j+1] == b[i]) j++;
        zfail[i] = j;
    }
    j = m+1;
    for(int i = m-1; i >= 1; --i) {
        while(j<=m&&b[j-1]!=b[i]) j = ffail[j];
        if(b[j-1] == b[i]) j--;
        ffail[i] = j;
    }
    j = 0;
    for(int i = 1; i <= n; ++i) {
        while(j&&a[i]!=b[j+1]) j = zfail[j];
        if(b[j+1] == a[i]) j++;
        dp[i][j] += 1;
    }
    j = m+1;
    for(int i = n; i >= 1; --i) {
        while(j<=m&&b[j-1]!=a[i]) j = ffail[j];
        if(b[j-1] == a[i]) j--;
        f[i][j] += 1;
    }

    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            dp[i][j] += dp[i-1][j];
            sumdp[i][j] += sumdp[i-1][j]+dp[i][j];
        }
    }
    for(int i = n; i >= 1; --i) {
        for(int j = m; j >= 1; --j) {
            f[i][j] += f[i+1][j];
            sumf[i][j] += sumf[i+1][j]+f[i][j];
        }
    }
}

void init2() {
     for(int i = 0; i <= n+1; ++i)
        for(int j = 0; j <= m+1; ++j)
            dp2[i][j] = 0,f2[i][j] = 0;
    for(int i = 0; i <= n+1; ++i)
        dp2[i][0] = 1,f2[i][m+1] = 1;

    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            if(a[i] == b[j] && dp2[i-1][j-1])
                dp2[i][j] = 1;
        }
    }
     for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
                dp2[i][j] += dp2[i-1][j];
        }
    }
    for(int i = n; i >= 1; --i) {
        for(int j = m; j >= 1; --j) {
            if(a[i] == b[j] && f2[i+1][j+1])
                f2[i][j] = 1;
        }
    }
      for(int i = n; i >= 1; --i) {
        for(int j = m; j >= 1; --j) {
          f2[i][j] += f2[i+1][j];
        }
    }
}

int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d%d%s%s",&n,&m,&k,a+1,b+1);
        init();
        init2();
        while(k--) {
            int L,R;
            scanf("%d%d",&L,&R);
            printf("%lld\n",solve(L,R));
        }
    }
    return 0;
}