FOJ 2200 Cleaning (ring DP)

Problem 2200 Cleaning problem Description

n people in a circle in the discussion of the big sweep of things, need to elect the K-man. But everyone has a contradiction with his distance of 2 people, so the distance of any two of the K-man cannot be 2, and they want to know how many methods there are. Input

The first line contains a number T (t<=100), which indicates the number of test data.

Next, there are two numbers n,k,n for each line, and K indicates the number of people needed (1<=n<=1000,1<=k<=n). Output

The output satisfies the test instructions scheme number, the scheme number is very large, so please output the program number mod 1,000,000,007 result. Sample Input 3 sample Output 416

First of all, consider the linear, the ring is actually a deformation of the line, Dp[i][j] means the former I personally take J personal situation, then dp[i][j]=dp[i-1][j]+dp[i-4][j-2]+dp[i-3][j-1]; DP[I-1][J] indicates that I am not selected//dp[i-4][j-2] to select I and i-1 individuals, do not select the I-2 individual//dp[i-3][j-1] to select the first person, do not select I-1 and i-2 next change line into a ring, is 1th and nth together (1 When 1,n is not selected, it is dp[n-2][k]; (2) When 1,n is selected, for Dp[n-6][k-2] (3) Elected 1 does not select N, this situation to consider the 2nd choice, the total is Dp[n-6][k-2]+dp[n-5][k-1] (4) elected n 1 O'Clock, and (3) is the same;

AC Code:

# include <stdio.h>
# include <string.h>
typedef long Long int ll;
const int mod=1000000007;
int dp[1010][1010];
int a[10][10];
int main () {
	int i, j, K, N, T;
	memset (DP, 0, sizeof (DP));
	Dp[1][0]=dp[1][1]=dp[0][0]=1;
	Dp[2][0]=1;dp[2][1]=2;dp[2][2]=1;
	dp[3][0]=1;dp[3][1]=3;dp[3][2]=2;
	for (i=4; i<=1000; i++) {
		dp[i][0]=1;
		dp[i][1]=i;
		for (j=2; j<=i; J + +) {
			dp[i][j]= ((Dp[i-1][j]+dp[i-4][j-2])%mod+dp[i-3][j-1])%mod;}
	}
    A[1][1]=1;a[2][1]=1;a[2][2]=1;
    a[3][1]=3;a[3][2]=3;a[4][1]=4;
    a[4][2]=4;a[5][1]=1;a[5][2]=5;
	scanf ("%d", &t);
	while (t--) {
		scanf ("%d%d", &n, &k);
		if (n<6) {
			printf ("%d\n", A[n][k]);
			Continue;
		}
		ll Ans=0;
		ANS=ANS+DP[N-2][K];
		ANS=ANS+DP[N-6][K-2];
		ANS=ANS+DP[N-6][K-2]+DP[N-5][K-1];
		ANS=ANS+DP[N-6][K-2]+DP[N-5][K-1];
		printf ("%d\n", (int) (ANS%MOD));
	}
	return 0;
}