Algorithm
Problem link- https://www.spoj.com/problems/CHAIR/
CHAIR - Chairs
N chairs are placed in a circle.
There will be K attendants to a very important meeting.
However, the attendants do not like each other, so they do not want to sit beside each other in the meeting.
As the host of this important meeting, you want to find out how many ways there are to choose K chairs such that none of them are adjacent to each other.
Input
The first line of the input is an integer N (4<=N<=1000), which denotes the number of chairs.
The next line is an integer K (1<=K<=N), which denotes the number of attendants to the meeting.
Output
Output the total number of ways to choose K chairs from N chairs such that none of the chairs are adjacent.
Since the answer can get very large, output the answer modulo 1,000,000,003.
Example
Input: 4 2 Output: 2
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000003LL
#define EPS 1e-9
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
const int MAXN = 1e5+5;
int N, K;
int dp[2][1001][501];
ll solve(int fi, int n, int k){
if(k == K) return 1;
if(n > N) return 0;
if(n == N && fi) return 0;
if(dp[fi][n][k] != -1) return dp[fi][n][k];
ll res = 0;
if(n == 1) res += solve(1, n + 2, k + 1);
else res += solve(fi, n + 2, k + 1);
res += solve(fi, n + 1, k);
res %= MOD;
return dp[fi][n][k] = res;
}
int main(){
// io;
// cin >> N >> K;
memset(dp, -1, sizeof dp);
scanf("%d %d", &N, &K);
if(K > N/2){
printf("0\n");
// cout << 0 << endl;
}
else{
int res = solve(0, 1, 0);
// cout << res << endl;
printf("%d\n", res);
}
return 0;
}Input
2
Output
Demonstration
SPOJ Solution-Chairs-Solution in C, C++, Java, Python