# Cards Permutation

### Problem Statement :

```Alice was given the  integers from  to . She wrote all possible permutations in increasing lexicographical order, and wrote each permutation in a new line. For example, for , there are  possible permutations:

She then chose one permutation among them as her favorite permutation.

After some time, she forgot some elements of her favorite permutation. Nevertheless, she still tried to write down its elements. She wrote a  in every position where she forgot the true value.

She wants to know the sum of the line numbers of the permutations which could possibly be her favorite permutation, i.e., permutations which can be obtained by replacing the s. Can you help her out?

Since the sum can be large, find it modulo .

Input Format

The first line contains a single integer .

The next line contains  space-separated integers  denoting Alice's favorite permutation with some positions replaced by .

Constraints

The positive values appearing in  are distinct.

For ~33% of the total points,
Output Format

Print a single line containing a single integer denoting the sum of the line numbers of the permutations which could possibly be Alice's favorite permutation.```

### Solution :

```                            ```Solution in C :

In   C++  :

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

const ll mod = 1000*1000*1000+7;

vector<ll> fact;
int N, K;
vector<int> P;

struct BIT {
vector<int> tree;
void init() {
tree = vector<int>(4*N, 0);
}
void upd(int idx, int val, int l,
int r, int n) {
if(idx < l || r < idx) return;
if(l == r) {
tree[n] = val;
return;
}
int m = (l + r)>>1;
upd(idx, val, l, m, 2*n);
upd(idx, val, m + 1, r, 2*n + 1);
tree[n] = tree[2*n] + tree[2*n + 1];
}
int quer(int a, int b, int l, int r, int n) {
if(b < l || r < a) return 0;
if(a <= l && r <= b) return tree[n];
int m = (l + r)>>1;
int L = quer(a, b, l, m, 2*n);
int R = quer(a, b, m + 1, r, 2*n + 1);
return L + R;
}
} bit, sub;

int main() {
fact.resize(500010);
fact = 1;
for(int i = 1; i < 500010; i++) {
fact[i] = fact[i - 1] * i % mod;
}

scanf("%d", &N);
K = 0;
P.resize(N);
sub.init();
for(int i = 0; i < N; i++) {
scanf("%d", &P[i]);
P[i]--;
if(P[i] == -1) K++;
sub.upd(i, 1, 0, N - 1, 1);
}

ll sum = 1LL * N * (N - 1) / 2;
for(int i = 0; i < N; i++) {
if(P[i] != -1) sum -= P[i], sub.upd(P[i], 0, 0, N - 1, 1);
}
sum = (sum % mod + mod) % mod;

bit.init();
ll ans = 0;
ll tmp = 0;
int cnt = 0;
for(int i = 0; i < N; i++) {
if(P[i] == -1) {
ll a = sum * fact[K - 1] % mod;
ll b = tmp * fact[K - 1] % mod;
ll c = K < 2? 0 : (1LL * K * (K - 1) / 2 % mod)
* fact[K - 2] % mod * cnt % mod;

ans += (a - b - c) * fact[N - i - 1] % mod,
ans = (ans % mod + mod) % mod;

cnt++;
}
else {
bit.upd(P[i], 1, 0, N - 1, 1);
tmp += sub.quer(P[i] + 1, N - 1, 0, N - 1, 1);
tmp %= mod;

ll a = fact[K] * P[i] % mod;
ll b = fact[K] * bit.quer(0, P[i] - 1, 0, N - 1, 1) % mod;
ll c = K == 0? 0 : fact[K - 1] * sub.quer(
0, P[i] - 1, 0, N - 1, 1) % mod * cnt % mod;

ans += (a - b - c) * fact[N - 1 - i] % mod,
ans = (ans % mod + mod) % mod;
}
}
cout << (ans + fact[K]) % mod;
}

In   C :

#include<stdio.h>
int n, a, pos,
mod = 1e9 + 7, occ,
les, grt, st,
lst, gst, bitree;
{
while( idx <= n )
{
bitree[idx] += val;
idx += ( idx & -idx );
}
}
int get(int idx)
{
int ans = 0;
while( idx > 0 )
{
ans += bitree[idx];
idx -= ( idx & -idx );
}
return ans;
}
long long fact, factsumfr, ans = 0;
long long pwr(long long a, long long b)
{
if( b == 0 )
{
return 1ll;
}
long long temp = pwr(a, b/2);
temp = ( temp * temp ) % mod;
if( b & 1 )
{
temp = ( temp * a ) % mod;
}
return temp;
}
long long inv(long long a)
{
return pwr(a, mod-2);
}
int main()
{
int i;
scanf("%d", &n);
for( i = 1 ; i <= n ; i++ )
{
scanf("%d", &a[i]);
pos[a[i]] = i;
if(a[i])
{
st[a[i]] = 1;
}
if(a[i])
{
occ[i] = 1;
}
}
fact = 1;
for( i = 1 ; i <= n ; i++ )
{
les[i] = les[i-1] + occ[i], lst[i] =
lst[i-1] + st[i], fact[i] = ( fact[i-1] * i ) % mod;
}
for( i = n ; i >= 1 ; i-- )
{
grt[i] = grt[i+1] + occ[i], gst[i] = gst[i+1] + st[i];
}
int k = les[n];
long long faci = fact[n-k],
faci1 = fact[n-k-1], sumfrfr = 0;
for( i = 1 ; i <= n ; i++ )
{
if( a[i] == 0 )
{
sumfrfr = ( sumfrfr + ( ( fact[n-i] * (
n - i - grt[i+1] ) ) % mod * inv(n-k-1) ) % mod ) % mod;
factsumfr[i] = (
factsumfr[i-1] + fact[n-i] ) % mod;
}
else
{
factsumfr[i] = factsumfr[i-1];
}
}
for( i = 1 ; i <= n ; i++ )
{
long long inc = 0;
if( st[i] == 0 )
{
inc += ( inc + ( ( sumfrfr * ( i - 1 - lst[i] )
) % mod * fact[n-k-1] ) % mod ) % mod;
}
else
{
inc = ( inc + ( ( ( n - i + 1 - gst[i] )
* factsumfr[pos[i]] ) % mod * fact[n-k-1] )
% mod ) % mod;
inc = ( inc + ( ( ( ( ( i - lst[i] )
* fact[n-pos[i]] ) % mod * fact[n-k] ) % mod * (
n - pos[i] + 1 - grt[pos[i]] ) ) % mod
* inv(n-k) ) % mod ) % mod;
int inv1 = get(n) - get(pos[i]);
inc = ( inc + ( ( fact[n-pos[i]]
* fact[n-k] ) % mod * inv1 ) % mod ) % mod;
}
ans = ( ans + inc ) % mod;
}
ans = ( ans + fact[n-k] ) % mod;
printf("%lld\n", ans);
return 0;
}

In   Python3  :

import math
import os
import random
import re
import sys
from bisect import bisect , insort
N =  pow(10,9)+7

def update(bit, i, v):
n = len(bit)
while i < n :
bit[i]+=v
i+=i&(-i)

def getsum(bit, i):
s=0
while i>0 :
s+=bit[i]
i -= i&(-i)
return s

def getnP(P,fixed):
n = len(P)
m = max(P)
nI=
for i in range(2,n+1):
nI.append(nI[-1] + fixed[i-2] )
bit = [0 for i in range(m+1)]

nP = [0 for i in range(n)]
for i in range(n-1,-1,-1):
if P[i] > 0 :
nP[i] = nI[P[i]-1] - getsum(bit, P[i]-1)
update(bit, P[i], 1)
else :
nP[i] = -1
nP = 0

return nP

def solve(P):
n = len(P)
fixed = n*

for v in P:
if v > 0 :
fixed[v-1] = 1

idZ   = [i for i in range(n) if P[i]==0 ]
idNZ  = [i for i in range(n) if P[i]>0 ]
vP    = [i for i in range(1,n+1) if not fixed[i-1]]
nz    = len(idZ)
nV,nZ,f = ,,

for i in range(1,n):
f.append( f[-1]*i %N )
nV.append(nV[-1]+(not fixed[i-1]))
nZ.append(nZ[-1]+(not P[i-1]))
f.append( f[-1]*n %N)

nP = getnP(P,fixed)

Tnz = sum( ( P[i] - 1 - nP[i] ) * f[n-i-1] for i in idNZ  )
Tz  = sum( ( i - nZ[i] ) * f[n-i-1] for i in idZ )
S  = f[nz] * ( Tnz - Tz + 1) %N

if nz > 0 :
svP = sum(j-1 for j in vP)
snPV = 
for i in range(1,n):
snPV.append( snPV[-1] + nV[P[i-1]-1]*(P[i-1]>0) )
Tnz = sum( nZ[i] * nV[P[i]-1] * f[n-i-1] for i in idNZ )
Tz  = sum( ( svP + snPV[i] ) * f[n-i-1] for i in idZ )
S += f[nz-1] * ( Tz - Tnz ) %N

if nz > 1 :
Tz = sum( nZ[i] * f[n-i-1] for i in idZ ) * sum( nV[l-1] for l in vP if l>1 )
S -= f[nz-2] * Tz %N

return S%N
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

n = int(input())

a = list(map(int, input().rstrip().split()))

result = solve(a)

fptr.write(str(result) + '\n')

fptr.close()```
```

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