Codeforces 1188C Array Beauty dp(看题解）

超、凢脫俗 2021-12-17 00:21 143阅读 0赞

[Array Beauty][]

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;
    }
    
    /*
    */

转载于:https://www.cnblogs.com/CJLHY/p/11142571.html

[Array Beauty]: https://codeforces.com/problemset/problem/1188/C