Codefroces 213E. Two Permutations

E. Two Permutations

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Rubik is very keen on number permutations.

A permutation a with length n is a sequence, consisting of n different numbers from 1 to n. Element number i (1 ≤ i ≤ n) of this permutation will be denoted as a_i.

Furik decided to make a present to Rubik and came up with a new problem on permutations. Furik tells Rubik two number permutations: permutation a with length n and permutation b with length m. Rubik must give an answer to the problem: how many distinct integers d exist, such that sequence c (c₁ = a₁ + d, c₂ = a₂ + d, ..., c_n = a_n + d) of length n is a subsequence of b.

Sequence a is a subsequence of sequence b, if there are such indices i₁, i₂, ..., i_n (1 ≤ i₁ < i₂ < ... < i_n ≤ m), that a₁ = b_i1, a₂ = b_i2, ..., a_n = b_in, where n is the length of sequence a, and m is the length of sequence b.

You are given permutations a and b, help Rubik solve the given problem.

Input

The first line contains two integers n and m (1 ≤ n ≤ m ≤ 200000) — the sizes of the given permutations. The second line contains n distinct integers — permutation a, the third line contains m distinct integers — permutation b. Numbers on the lines are separated by spaces.

Output

On a single line print the answer to the problem.

Examples

Input

1 1
1
1

Output

1

Input

1 2
1
2 1

Output

2

Input

3 3
2 3 1
1 2 3

Output

0

【题解】

由于a和b都是排列，其实就是找b中i...i + n - 1的相对位置与a中1...n的相对位置是否相同

康托展开显然是不好做的（反正我不会）

于是我们可以hash，拿线段树做即可

线段树下标是排列位置，线段数内的值是该位置的值

先把1...n的位置放进去，然后查行不行

然后把1拿出来，把n + 1的位置放进去，看看行不行

（注意由于每个数字都增加1，因此hash值应增加B^0 + B^1 + B^2.. + B^n，前缀和处理即可）

以此类推

因为没有膜够所以WA了好几发

具体看代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
    if(c == '-')x = -x;
}

const long long MAXN = 1000000 + 10;
const long long MOD1 = 19260817;
const long long MOD2 = 1000000007;
const long long B = 232;
const long long MOD = 1000000007;

long long data[2][MAXN], ans, sum1, sum2, sum[MAXN], val1, val2, bit1[MAXN], bit2[MAXN], cnt[MAXN << 1], size1[MAXN], size2[MAXN], n, m;

//在权值p这个位置，[x == 1]增加k,[x == -1]减小k 
void modify(long long p, long long k, long long x, long long o = 1, long long l = 1, long long r = m)
{
    if(l == r && p == l)
    {
        data[0][o] += k * x % MOD1;
        data[1][o] += k * x % MOD2;
        size1[o] += x, size2[o] += x;
        return;
    }
    long long mid = (l + r) >> 1;
    if(mid >= p) modify(p, k, x, o << 1, l, mid);
    else modify(p, k, x, o << 1 | 1, mid + 1, r);
    data[0][o] = ((data[0][o << 1 | 1] * bit1[size1[o << 1]]) % MOD1 + data[0][o << 1]) % MOD1;
    data[1][o] = ((data[1][o << 1 | 1] * bit2[size2[o << 1]]) % MOD2 + data[1][o << 1]) % MOD2;
    size1[o] = size1[o << 1] + size1[o << 1 | 1];
    size2[o] = size2[o << 1] + size2[o << 1 | 1];
    return;
}

int main()
{
//    freopen("data.txt", "r", stdin);
    read(n) ,read(m);
    bit1[0] = 1;bit2[0] = 1;
    for(register long long i = 1;i <= m;++ i)
        bit1[i] = (bit1[i - 1] * B)%MOD1, bit2[i] = (bit2[i - 1] * B)%MOD2;
    for(register long long i = 1;i <= n;++ i)
    {
        long long tmp;
        read(tmp);
        val1 += tmp * bit1[i - 1] % MOD1;
        val1 %= MOD1;
        val2 += tmp * bit2[i - 1] % MOD2;
        val2 %= MOD2;
        sum1 += bit1[i - 1];sum1 %= MOD1;
        sum2 += bit2[i - 1];sum2 %= MOD2;
    }
    for(register long long i = 1;i <= m;++ i)
    {
        long long tmp;
        read(tmp);
        cnt[tmp] = i;
    }
    for(register long long i = 1;i <= m;++ i)
    {
        modify(cnt[i], i, 1);
        if(i >= n)
        {
            if(data[1][1] == (val2 + (sum2 * (i - n))%MOD2)%MOD2 && data[0][1] == (val1 + (sum1 * (i - n))%MOD1)%MOD1)
                ++ ans;
            modify(cnt[i - n + 1], i - n + 1, -1);
        }
    } 
    printf("%d", ans);
    return 0;
}