DP专题练习3

安全序列

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
const int N = 1e6 + 10,mod = 1e9 + 7;
int n, k;
int dp[N];
void solve(){
	cin >> n >> k;
	dp[1] = 2;
	for(int i = 2; i <= n ; i ++){
		if(i-k-1>=1) dp[i] = (dp[i-1] + dp[i-k-1])%mod;
		else dp[i] = (dp[i-1] + 1) % mod;//如果i-k-1<1,如果选可以让前面全为0,最后加1
	}
	cout << dp[n];
}
signed main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int _ = 1;
	//cin >> _;
	while(_--) solve();
	return 0;
}