Array Beauty CodeForces - 1189F (dp,好题)

大意: 定义$n$元素序列$a$的美丽度为 $\min\limits_{1\le i<j\le n}|a_i-a_j|$.

给定序列$a$, 求$a$的所有长为$k$的子序列的美丽度之和.

记 长为$k$的相邻元素距离都$\ge x$的子序列个数 为$f(x)$.

那么答案就为$\sum\limits_{x=1}^{\frac{1e5}{k-1}} f(x)$.

$f(x)$很容易可以$O(nk)$的$dp$求出, 总复杂度就为$O(1e5n)$.

#include <iostream>
#include <algorithm>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
const int N = 1e3+10, P = 998244353;
int n, k, a[N];
int dp[N][N];
int solve(int x) {
    dp[0][0] = 1;
    int ans = 0, now = 0;
    REP(i,1,n) {
        while (a[i]-a[now+1]>=x) ++now;
        dp[i][0] = dp[i-1][0];
        REP(j,1,k) dp[i][j] = (dp[i-1][j]+dp[now][j-1])%P;
    }
    return dp[n][k];
}

int main() {
    scanf("%d%d", &n, &k);
    REP(i,1,n) scanf("%d", a+i);
    sort(a+1,a+1+n);
    int ans = 0;
    REP(i,1,1e5/(k-1)) ans = (ans+solve(i))%P;
    printf("%d\n", ans);
}