【Complete Backpack with Restrictions】Educational Codeforces Round 133 (Rated for Div. 2) D. Chip Move

题意：
给定 $n$ 和 $k$,初始步长为 $k$ ,每次可以走 $k$ multiples of steps,next time go $(k+1)$multiples of steps,以此类推,问从 $0$ 开始到达 $[1,n]$ The number of scenarios per point.

题解：
When only taking double the number of steps each time,$\frac{(step+k)\times (step-k+1)}{2}\le n$,So at most just go $O(\sqrt{n})$次.

The problem is transformed into a complete knapsack problem with constraints,走第 $i$ 次前,must have gone $1$ 到 $i-1$次

$f[i][j]$ 表示走了 $i$次,第 $i$ Arrive at the point after the walk $j$ 的方案数
第 $i$ The number of steps taken is $(k+i-1)$ 的倍数

The template for a full backpack without restrictions is ：

for (int i = 1; i <= n; ++i) f[i] = 0;
f[0] = 1;

for (int step = k; (start = (step + k) * (step - k + 1) / 2) <= n; ++step)
	for (int j = 0; j <= n; ++j)
		if (j >= step) f[i][j] += f[i - 1][j - step];

But this question is limited to ：走第 $i$ 次前,must have gone $1$ 到 $i-1$次
Every time you need to force a walk $i-1$ 次,如下：

for (int i = 1; i <= n; ++i) f[0][i] = 0;
f[0][0] = 1;
for (int step = k; (step + k) * (step - k + 1) / 2 <= n; ++step) {
    
	for (int j = 0; j <= n; ++j)
		if (j >= step) f[i][j] = f[i - 1][j - step];
		else f[i][j] = 0;
	for (int j = 0; j <= n; ++j)
		if (j >= step) f[i][j] += f[i - 1][j - step];
	for (int j = 1; j <= n; ++j) ans[j] += f[j];
}

Optimize the scrolling array：

for (int i = 1; i <= n; ++i) f[i] = 0;
f[0] = 1;

for (int step = k; (step + k) * (step - k + 1) / 2 <= n; ++step) {
    
	for (int j = n; j >= 0; --j)
		if (j >= step) f[j] = f[j - step];
		else f[j] = 0;
		
	for (int j = 0; j <= n; ++j)
		if (j >= step) f[j] += f[j - step];
	for (int j = 1; j <= n; ++j) ans[j] += f[j];
}

代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 200010;
const int mod = 998244353;

int dp[2][N];
int f[N];
int ans[N];
int n, k;

void add(int& a, int b) {
    
	a += b;
	if (a >= mod) a -= mod;
}

void solve() {
    
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; ++i) f[i] = 0;
	f[0] = 1;
	
	for (int step = k, start; (start = (step + k) * (step - k + 1) / 2) <= n; ++step) {
    
		for (int j = n; j >= 0; --j)
			if (j >= start) f[j] = f[j - step];
			else f[j] = 0;
			
		for (int j = start; j <= n; ++j)
			if (j >= start) add(f[j], f[j - step]);
		for (int j = 1; j <= n; ++j) add(ans[j], f[j]);
	}
	for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]); 
}

int main()
{
    
	int T = 1;
// scanf("%d", &T);
	for (int i = 1; i <= T; ++i) {
    
		solve();
	}
	return 0;
}

总结：
Avoid jumping thinking,After constructing the transition equation from the naive 2D state,Then optimize the space and time according to the actual situation