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. Subtask 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[0] = 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[300100], pos[300100],
mod = 1e9 + 7, occ[300100],
les[300100], grt[300100], st[300100],
lst[300100], gst[300100], bitree[300050];
void add(int idx, int val)
{
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[300100], factsumfr[300100], 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[0] = 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;
add(pos[i], 1);
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=[0]
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] = 0
return nP
def solve(P):
n = len(P)
fixed = n*[0]
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 = [0],[0],[1]
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 = [0]
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()
View More Similar Problems
Waiter
You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the
View Solution →Queue using Two Stacks
A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que
View Solution →Castle on the Grid
You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):
View Solution →Down to Zero II
You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.
View Solution →Truck Tour
Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr
View Solution →Queries with Fixed Length
Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon
View Solution →