[Agc028B] Removing Blocks_ permutations

Removing Blocks

Topic links : https://atcoder.jp/contests/agc028/tasks/agc028_b

Data range : Slightly.

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Problem solution :

The first step of this problem is very routine, is seeking separately for each $ a_i $.

So how should each $ a_i $ demand for it?

Consider deleted $ j $, when there is $ a_i $ contributions, how many kinds of programs.

In this case, the number needed to ensure that all $ i \ sim j $ the middle have been deleted.

We consider permutations when the broad sense is to put anyone indifferent.

Might put the number of those should appear after the first $ j $ into them, so the time to put the $ j $ is there is only one solution.

Program number is the $ \ frac {n!} {Len _ {(j \ rightarrow i)}} $.

This thing is $ O (n ^ 2) $, the prefix and optimization look into $ O (n) $ a.

Code :

#include <bits/stdc++.h>

#define N 300010 

using namespace std;

typedef long long ll;

const int mod = 1000000007 ;

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
    int x = 0, f = 1;
    char c = nc();
    while (c < 48) {
        if (c == '-')
            f = -1;
        c = nc();
    }
    while (c > 47) {
        x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
    }
    return x * f;
}

int a[N], fac[N], fac2[N], bfr[N];

int qpow(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) {
            ans = (ll)ans * x % mod;
        }
        y >>= 1;
        x = (ll)x * x % mod;
    }
    return ans;
}

int main() {
    int n = rd();
    for (int i = 1; i <= n; i ++ ) {
        a[i] = rd();
    }

    // init
    fac[0] = 1;
    for (int i = 1; i <= n; i ++ ) {
        fac[i] = (ll)fac[i - 1] * i % mod;
    }
    for (int i = 1; i <= n; i ++ ) {
        fac2[i] = (ll)fac[n] * qpow(i, mod - 2) % mod;
        bfr[i] = (bfr[i - 1] + fac2[i]) % mod;
    }

    // for (int i = 1; i <= n; i ++ ) {
    //     printf("%d ", fac[i]);
    // }
    // puts("");
    // for (int i = 1; i <= n; i ++ ) {
    //     printf("%d ", fac2[i]);
    // }
    // puts("");

    int ans = 0;
    for (int i = 1; i <= n; i ++ ) {
        ans = (ans + (ll)a[i] * (
            (((ll)bfr[i] + bfr[n - i + 1]) % mod + mod - fac[n]) % mod
        )) % mod;
    }

    cout << ans << endl ;
    return 0;
}