The Two Types of Contiguous Segments

CCPC 2021 Online H - Subpermutation defines the full-permutation of $n$: concatenate all permutations of $1$ through $n$ in lexicographic order to obtain the sequence $p_n$.

Given $n,m$, count the contiguous subsequences of $p_n$ of length $m$ that are themselves permutations of $1$ through $m$. Answers for multiple test cases are taken modulo $10^9+7$, with:

Each original permutation has length $n$, and the segment length $m$ does not exceed $n$. A segment can therefore meet at most two original permutations, and all cases split completely into two types:

• The segment lies entirely within one permutation;
• The segment takes a nonempty suffix of one permutation followed by a nonempty prefix of the next.

These two types are disjoint, so count them separately and add the results.

Contributions Within One Permutation

A contiguous segment of length $m$ is a permutation of $1$ through $m$ if and only if those $m$ values occupy $m$ consecutive positions in the current permutation.

First choose the starting position of this block, giving $n-m+1$ choices. The values $1$ through $m$ can be arranged arbitrarily inside the block, and the remaining $n-m$ values can also be arranged arbitrarily. The total number of valid segments lying entirely within a single permutation is therefore:

This counts across all $n!$ permutations together. Each permutation satisfying the condition contributes exactly one valid starting position, with no duplication.

Crossing a Permutation Boundary

Fix the length taken from the end of the preceding permutation as $k$, where:

Let $d=m-k$, and split the preceding permutation $P$ by position into:

Here, $|A|=d$, $|B|=n-m$, and $|C|=k$. The segment crossing the boundary consists of $C$ followed by the length-$d$ prefix of the successor permutation $Q$.

First, require only that $A$ and $C$ together contain exactly the values $1$ through $m$. Then:

• The values $1$ through $m$ can occupy the $m$ positions of $A,C$ in any order, giving $m!$ choices;
• The values $m+1$ through $n$ can occupy the middle part $B$ in any order, giving $(n-m)!$ choices.

The number of candidate permutations is:

We still need to determine whether prefix $A$ stays unchanged when moving from $P$ to its lexicographic successor $Q$.

The Subtraction Term from the Longest Decreasing Suffix

To find a permutation’s lexicographic successor, locate its longest decreasing suffix, swap the pivot immediately before that suffix, and put the suffix into its smallest, ascending order.

If $B+C$ is not decreasing, the pivot must lie within $B+C$, so the length-$d$ prefix $A$ remains unchanged. The segment crossing the boundary is then $C+A$, containing each value from $1$ through $m$ exactly once, and is valid.

Conversely, if all of $B+C$ is decreasing, the lexicographic-successor operation crosses the boundary between $A$ and $B+C$. The first $d$ entries of the successor are no longer exactly the original $A$: an element originally to the right of that boundary moves into the prefix. If $P$ is already the last permutation, it has no successor at all. In either case, $C$ and the new prefix can no longer form exactly a permutation of $1$ through $m$.

We therefore only need to subtract the cases where all of $B+C$ is decreasing from the candidate count.

Every value in $B$ is greater than $m$, while every value in $C$ is at most $m$, so their boundary is already decreasing. For all of $B+C$ to be decreasing, it is enough for $B$ and $C$ each to be in descending order:

• Choose the $k$ values of $C$ from $1$ through $m$, giving $\binom{m}{k}$ choices. Once chosen, their descending order in $C$ is unique;
• Arrange the remaining $d$ values arbitrarily in $A$, giving $d!$ choices;
• The descending order of $B$ is unique.

The subtraction term is therefore:

For fixed $k$, the number of valid segments crossing a boundary is:

Summing over all $k$:

Combining the Counts into a Closed Form

Add the contributions from within permutations and across their boundaries:

Ultimately, we need only factorials, inverse factorials, and prefix sums of inverse factorials.

Let:

In the implementation, fac[i], ifac[i], and sum[i] maintain $i!$, $(i!)^{-1}$, and $S_i$, respectively. Each query directly evaluates:

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return (1ll * n * fac[n - m] % mod - sum[m - 1] + mod) % mod * fac[m] % mod;

All divisions here use modular inverses modulo $10^9+7$. The preprocessing range is below the modulus, so every factorial has an inverse.

When $m=1$, the sum is empty and sum[0]=0. The formula gives $n(n-1)!=n!$, exactly the number of occurrences of 1 in the full sequence. When $n=m$, fac[n-m]=fac[0]=1, so the same formula works without an additional branch.

Complexity and Implementation Boundaries

Precompute factorials, inverse factorials, and prefix sums up to $10^6$. This takes $O(10^6+\log(10^9+7))$ time and $O(10^6)$ space. After preprocessing, each query performs only a constant number of modular operations, taking $O(1)$ per query and $O(T)$ for all queries.

Three implementation boundaries must also remain consistent:

• Set fac[0]=1, so $0!$ is available directly when $n=m$;
• Start the inverse-factorial prefix sum at ifac[1], so the subtraction term is empty when $m=1$;
• For modular subtraction, add mod before taking the remainder to avoid a negative intermediate result.

Final Accepted Code

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int mod = 1e9 + 7, maxn = 1e6 + 5;

ll t, fac[maxn], ifac[maxn], sum[maxn];

ll qpow(ll a, ll b) {
    ll res = 1;
    a %= mod;
    while (b) {
        if (b & 1) {
            res = res * a % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

void init() {
    fac[0] = 1;
    for (int i = 1; i < maxn; ++i) {
        fac[i] = fac[i - 1] * i % mod;
    }
    ifac[maxn - 1] = qpow(fac[maxn - 1], mod - 2) % mod;
    for (int i = maxn - 2; i >= 0; --i) {
        ifac[i] = ifac[i + 1] * (i + 1) % mod;
    }
    for (int i = 1; i < maxn; ++i) {
        sum[i] = (sum[i - 1] + ifac[i]) % mod;
    }
}

ll ans(int n, int m) {
    return (1ll * n * fac[n - m] % mod - sum[m - 1] % mod + mod) % mod * fac[m] % mod;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    init();
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        cout << ans(n, m) << '\n';
    }
    return 0;
}