Dortmund Dilemma
Problem Statement :
Borussia Dortmund are a famous football ( soccer ) club from Germany. Apart from their fast-paced style of playing, the thing that makes them unique is the hard to pronounce names of their players ( błaszczykowski , papastathopoulos , großkreutz etc. ). The team's coach is your friend. He is in a dilemma as he can't decide how to make it easier to call the players by name, during practice sessions. So, you advise him to assign easy names to his players. A name is easy to him if 1. It consists of only one word. 2. It consists of only lowercase english letters. 3. Its length is exactly N. 4. It contains exactly K different letters from the 26 letters of English alphabet. 5. At least one of its proper prefixes matches with its proper suffix of same length. Given, N and K you have to tell him the number of easy names he can choose from modulo (10^9+7). Note : A prefix P of a name W is proper if, P!=W. Similarly, a suffix S of a name W is proper if, S!=W. Input Format The first line of the input will contain T ( the number of testcases ). Each of the next lines will contain space separated integers N and K. Output Format For each testcase, output the number of ways the coach can assign names to his players modulo (10^9+9). Constraints 1 <= T <= 10^5 1 <= N <= 10^5 1 <= K <= 26
Solution :
Solution in C :
In C++ :
#include <bits/stdc++.h>
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define RS(X) scanf("%s", (X))
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
typedef long long LL;
using namespace std;
const int SIZE = 1e5+5;
const int MOD = 1e9+9;
LL C[27][27],an[27][SIZE],tmp[27][SIZE],pp[27][SIZE],tmp2[27][SIZE],ha[27];
int main(){
REPP(i,1,27){
pp[i][0]=1;
REPP(j,1,SIZE)pp[i][j]=pp[i][j-1]*i%MOD;
}
REP(i,27){
C[i][0]=1;
REPP(j,1,i+1){
C[i][j]=C[i-1][j-1]+C[i-1][j];
}
}
REPP(i,1,27){
an[i][1]=i;
tmp[i][1]=i;
REPP(j,2,SIZE){
if(j%2==0)an[i][j]=tmp[i][j/2];
else an[i][j]=tmp[i][j/2]*i%MOD;
an[i][j]=pp[i][j]-an[i][j];
REPP(k,1,i){
an[i][j]=(an[i][j]-an[k][j]*C[i][k])%MOD;
}
if(an[i][j]<0)an[i][j]+=MOD;
tmp[i][j]=tmp[i][j-1]*i*i%MOD;
REPP(k,1,i+1){
tmp[i][j]=(tmp[i][j]+C[i][k]*an[k][j])%MOD;
}
}
}
CASET{
DRII(N,K);
if(N==1){
puts("0");
continue;
}
LL res=0;
if(N%2==0){
ha[1]=tmp[1][N/2];
REPP(i,2,K+1){
ha[i]=tmp[i][N/2];
REPP(j,1,i)ha[i]=(ha[i]-ha[j]*C[i][j])%MOD;
if(ha[i]<0)ha[i]+=MOD;
}
res=ha[K]*C[26][K]%MOD;
}
else{
ha[1]=tmp[1][N/2];
REPP(i,2,K+1){
ha[i]=tmp[i][N/2]*i%MOD;
REPP(j,1,i)ha[i]=(ha[i]-ha[j]*C[i][j])%MOD;
if(ha[i]<0)ha[i]+=MOD;
}
res=ha[K]*C[26][K]%MOD;
}
if(K==1)res=26;
if(res<0)res+=MOD;
printf("%lld\n",res);
}
return 0;
}
In Java :
import java.io.*;
import java.util.Arrays;
public class Solution {
static final long MOD = 1000000009;
static final int MAXK = 26;
static final int MAXN = 100000;
static long modPow(long x, long pow) {
long r = 1;
while (pow > 0) {
if (pow % 2 == 1) {
r = r * x % MOD;
}
pow /= 2;
x = x * x % MOD;
}
return r;
}
public static void solve(Input in, PrintWriter out) throws IOException {
long[][] c = new long[MAXK + 1][MAXK + 1];
for (int i = 0; i < c.length; ++i) {
for (int j = 1; j < i; ++j) {
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
}
c[i][0] = c[i][i] = 1;
}
long[][] d = new long[MAXK + 1][MAXN + 1];
for (int j = 1; j <= MAXK; ++j) {
d[j][0] = 1;
}
for (int i = 1; i <= MAXN; ++i) {
for (int j = 1; j <= MAXK; ++j) {
d[j][i] = (d[j][i - 1] * j + (i % 2 == 0 ? MOD - d[j][i / 2] : 0)) % MOD;
}
}
for (int i = 1; i <= MAXN; ++i) {
for (int j = 1; j <= MAXK; ++j) {
d[j][i] = (modPow(j, i) + MOD - d[j][i]) % MOD;
for (int j1 = 1; j1 < j; ++j1) {
d[j][i] = (d[j][i] + (MOD - c[j][j1]) * d[j1][i]) % MOD;
}
}
}
int tests = in.nextInt();
for (int test = 0; test < tests; ++test) {
int n = in.nextInt();
int k = in.nextInt();
out.println(d[k][n] * c[MAXK][k] % MOD);
}
}
public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
solve(new Input(new BufferedReader(new InputStreamReader(System.in))), out);
out.close();
}
static class Input {
BufferedReader in;
StringBuilder sb = new StringBuilder();
public Input(BufferedReader in) {
this.in = in;
}
public Input(String s) {
this.in = new BufferedReader(new StringReader(s));
}
public String next() throws IOException {
sb.setLength(0);
while (true) {
int c = in.read();
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
int c = in.read();
if (c == -1 || " \n\r\t".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#define M (1000000009)
long long NCR[27][27];
long long DP[100001][27];
int main() {
int T,N,K;
for(int n = 0; n <= 26; n++) {
NCR[n][0] = 1;
NCR[n][n] = 1;
for(int r = 1; r < n; r++) {
NCR[n][r] = (NCR[n-1][r-1] + NCR[n-1][r])%M;
}
}
for(int k = 1; k <= 26; k++) {
DP[0][k] = 1;
for(int n = 1; n <= 100000; n++) {
DP[n][k] = (k*DP[n-1][k])%M;
if(n%2 == 0) {
DP[n][k] -= DP[n/2][k];
DP[n][k] %= M;
if(DP[n][k] < 0) {
DP[n][k] += M;
}
}
}
}
for(int k = 1; k <= 26; k++) {
long long X = 1;
for(int n = 0; n <= 100000; n++) {
DP[n][k] = X - DP[n][k];
if(DP[n][k] < 0) {
DP[n][k] += M;
}
X = (X*k)%M;
}
}
for(int k = 1; k <= 26; k++) {
for(int n = 0; n <= 100000; n++) {
for(int j = 1; j < k; j++) {
DP[n][k] -= (NCR[k][j]*DP[n][j])%M;
if(DP[n][k] < 0) {
DP[n][k] += M;
}
}
}
}
scanf("%d",&T);
for(int i = 0; i < T; i++) {
scanf("%d",&N);
scanf("%d",&K);
printf("%lld\n", (NCR[26][K]*DP[N][K])%M);
}
return 0;
}
In Python3 :
MOD = 10 ** 9 + 9
N = 10 ** 5
K = 26
class Solver(object):
def __init__(self):
self.factorials = [1 for _ in range(26 + 1)]
for i in range(1, 26 + 1):
self.factorials[i] = self.factorials[i - 1] * i
self.F = [[0 for _ in range(K + 1)] for _ in range(N + 1)]
for j in range(1, K + 1):
self.F[1][j] = j
for i in range(2, N + 1):
if i % 2 == 0:
self.F[i][j] = self.F[i - 1][j] * j - self.F[i // 2][j]
else:
self.F[i][j] = self.F[i - 1][j] * j
self.F[i][j] %= MOD
self.G = [[0 for _ in range(K + 1)] for _ in range(N + 1)]
for j in range(1, K + 1):
total = j
for i in range(1, N + 1):
self.G[i][j] = total - self.F[i][j]
self.G[i][j] %= MOD
total *= j
total %= MOD
def solve(self, n, k):
P = 0
if k == 1:
P += self.G[n][k]
else:
for j in range(k, 0, -1):
P += (-1) ** (k - j) * self.G[n][j] * (
self.factorials[k] // (self.factorials[j] * self.factorials[k - j]))
P %= MOD
res = P * (self.factorials[26] // (self.factorials[k] * self.factorials[26 - k]))
return res % MOD
if __name__ == "__main__":
import sys
testcases = int(input())
solver = Solver()
for _ in range(testcases):
n, k = map(int, input().split())
print(solver.solve(n, k))
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