【計數】tower

看了題解完全覺得是sb題。。。那麼 = =作爲一個sb題都做不來的。。。

設dp[i] 爲前i 個的方案數，那麼枚舉所有能在上面放i 的積木j，就會對答案帶來dp[i - 1]的貢獻。

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

int read()
{
	int n = 0, sign = 1; char c = getchar();
	while(c < '0' || c > '9') {if(c == '-') sign = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { n = n*10 + c-'0'; c = getchar(); }
	return n * sign;
}

const int Mod = 998244353;

int N, D;
int a[700005];
int ans;

int main()
{
	freopen("tower.in", "r", stdin);
	freopen("tower.out", "w", stdout);
	
	N = read(), D = read();
	for (int i = 1; i <= N; ++i) a[i] = read();
	sort(a + 1, a + N + 1); ans = 1;
	
	for (int i = 2, j = 1; i <= N; ++i) {
		while (a[j] + D < a[i]) ++j;
		ans = 1ll * ans * (i - j + 1) % Mod;
	}
	
	printf("%d\n", ans);
	
	return 0;
}