Codeforces 1188C Array Beauty dp(看题解)
最关键的一点是, 可以枚举最小值, 对于每个最小值算出大于等于它的方案数, 这个复杂度是科学的因为 i * (k - 1) <= dif
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 1000 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;} mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n, k, a[N]; int dp[N][N]; int sum[N]; int way[100007]; int calc(int d) { int ans = 0; for(int i = 0; i <= k; i++) { for(int j = 0; j <= n; j++) { dp[i][j] = 0; } } dp[0][0] = 1; for(int i = 0; i < k; i++) { sum[0] = dp[i][0]; for(int j = 1; j <= n; j++) { sum[j] = (sum[j - 1] + dp[i][j]) % mod; } int p = 1; for(int j = 1; j <= n; j++) { while(a[j] - a[p] >= d) p++; dp[i + 1][j] = sum[p - 1]; } } for(int j = k; j <= n; j++) { add(ans, dp[k][j]); } return ans; } int main() { scanf("%d%d", &n, &k); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } sort(a + 1, a + 1 + n); int dif = a[n] - a[1]; int ans = 0; for(int i = 1; i * (k - 1) <= dif; i++) { way[i] = calc(i); } for(int i = 1; i * (k - 1) <= dif; i++) { sub(way[i], way[i + 1]); add(ans, 1LL * i * way[i] % mod); } printf("%d\n", ans); return 0; } /* */