E - Distance Sequence (prefix and optimized dp

E - Distance Sequence
思路：
$dp[i][j]$ 表示考虑前 $i$ 个数,最后一个数为 $j$ 时的方案数
显然 $dp[i][j]+=dp[i-1][l]$,$l$ 为前 $i-1$ 个数中以 $l$ End and with $j$ In the case of the joint method
$j-l\ge k$,即 $1\le l\le j-k$
$l-j\ge k$,即 $j+k\le l\le m$
Obviously a prefix and optimization can be maintained
（Note that special judgment is required $K=0$ 的情况
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define mem(x, d) memset(x, d, sizeof(x))
#define eps 1e-6
using namespace std;
const int maxn = 2e6 + 9;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m;
ll dp[1009][5009];

void work()
{
    
	int k;
	cin >> n >> m >> k;
	if(k == 0){
    
		ll ans = 1;
		for(int i = 1; i <= n; ++i)
			ans = ans * m % mod;
		cout << ans;
		return;
	}
	vector <ll> sum(m + 1, 0);
	for(int i = 1; i <= m; ++i) {
    
		dp[1][i] = 1;
		sum[i] = sum[i-1] + dp[1][i];
	}
	for(int i = 2; i <= n; ++i){
    
		vector <ll> now_sum(m + 1, 0);
		for(int j = 1; j <= m; ++j){
    
			if(j - k >= 1)
				(dp[i][j] += sum[j - k]) %= mod;
			//for(int l = 1; l <= j - k; ++l)
			// (dp[i][j] += dp[i-1][l]) %= mod;
			if(j + k <= m)
				(dp[i][j] += (sum[m] - sum[j+k-1] + mod) % mod) %= mod;
			//for(int l = j + k; l <= m; ++l)
			// (dp[i][j] += dp[i-1][l]) %= mod;
		}
		for(int j = 1; j <= m; ++j) now_sum[j] = (now_sum[j-1] + dp[i][j]) % mod;
		sum = now_sum;
	}
	cout << sum[m];
}

int main()
{
    
	ios::sync_with_stdio(0);
// int TT;cin>>TT;while(TT--)
	work();
	return 0;
}

改进 $dp$ The process does not need to be judged individually,But also think about it $k==0$ the situation,Then modify the transfer equation
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define mem(x, d) memset(x, d, sizeof(x))
#define eps 1e-6
using namespace std;
const int maxn = 2e6 + 9;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m, k;
ll dp[1009][5009];
 
void work()
{
    
	cin >> n >> m >> k;
	
	vector <ll> sum(m + 1, 0);
	for(int i = 1; i <= m; ++i) {
    
		dp[1][i] = 1;
		sum[i] = sum[i-1] + dp[1][i];
	}
	
	for(int i = 2; i <= n; ++i){
    
		vector <ll> now_sum(m + 1, 0);
		for(int j = 1; j <= m; ++j){
    
			int l = max(1ll, j - k + 1), r = min(m, j + k - 1);
			if(k){
    
				dp[i][j] = (sum[m] - (sum[r] - sum[l-1]) + mod) % mod;
			}
			else {
    
				dp[i][j] = sum[m];
			}
		}
		for(int j = 1; j <= m; ++j) now_sum[j] = (now_sum[j-1] + dp[i][j]) % mod;
		sum = now_sum;
	}
	cout << sum[m];
}

int main()
{
    
	ios::sync_with_stdio(0);
// int TT;cin>>TT;while(TT--)
	work();
	return 0;
}

想到 $k=0$ Then change the transfer equation,I can't do it for a chicken like me,Hence the following writing,It can be completely ignored $K$ 的值了,Just make sure every transfer is legal（也就是 $r>=l-1$）
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define mem(x, d) memset(x, d, sizeof(x))
#define eps 1e-6
using namespace std;
const int maxn = 2e6 + 9;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m, k;
ll dp[1009][5009];
// 考虑前i个数,最后一个数为j
void work()
{
    
	cin >> n >> m >> k;
	
	vector <ll> sum(m + 1, 0);
	for(int i = 1; i <= m; ++i) {
    // 单独一个数1-m都可以
		dp[1][i] = 1;
		sum[i] = sum[i-1] + dp[1][i];
	}
	
	for(int i = 2; i <= n; ++i){
    
		vector <ll> now_sum(m + 1, 0);
		for(int j = 1; j <= m; ++j){
    
			int l = max(1ll, j - k + 1), r = min(m, j + k - 1);// 不合法的区间
			dp[i][j] = sum[m];
			if(r >= l - 1) {
    // If the illegal interval exists, it needs to be deleted
				dp[i][j] = (dp[i][j] - (sum[r] - sum[l-1]) + mod) % mod;
			}
		}
		for(int j = 1; j <= m; ++j) now_sum[j] = (now_sum[j-1] + dp[i][j]) % mod;
		sum = now_sum;
	}
	cout << sum[m];
}

int main()
{
    
	ios::sync_with_stdio(0);
// int TT;cin>>TT;while(TT--)
	work();
	return 0;
}