《 Codeforces Round #750 (Div. 2)》

E：我们从1开始考虑最大的k，从右开始到左，dp一下就好。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double ld;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e8
#define IN_INF 0x3f3f3f3f
#define dbg(ax) cout << "now this num is " << ax << endl;
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

LL dp[N][500],sum[N];//dp[i][j] - 以i~n为起点的长度为j的最大和
int a[N];
void solve() {
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;++i) scanf("%d",&a[i]),sum[i] = sum[i - 1] + a[i];
    for(int i = 1;i <= n + 1;++i) {
        for(int j = 1;j < 500;++j) dp[i][j] = 0;
    }
    int ans = -1;
    for(int i = n;i >= 1;--i) {
        for(int j = 1;j < 500;++j) {
            if(i + j - 1 > n) continue;
            LL ma = sum[i + j - 1] - sum[i - 1];
            if(dp[i + j][j - 1] > ma || j == 1) {
                dp[i][j] = ma;
                ans = max(ans,j);
            }
            dp[i][j] = max(dp[i + 1][j],dp[i][j]);
        }
    }
    printf("%d\n",ans);

}
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        solve();
    }
    //system("pause");
    return 0;
}

F1,F2做法一样：

我们其实只需要考虑对于i，能存入的异或值j，里<它的一个就行。

所以对于每个a[i]，我们维护它覆盖的最大区间，也就是一个比他大的异或值放入就够了，然后就减小。

这样复杂度max_a[i] *max_a[i]。

注意这里G[i]要清空，才能不重复计算。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 1e6 + 5;
const int M = 2e4 + 5;
const double eps = 1e-6;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 0x3f3f3f
#define dbg(ax) cout << "now this num is " << ax << endl;
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int a[N],r[1 << 13],up = (1 << 13) - 1;
bool vis[1 << 13];
vector<int> G[1 << 13];
void solve() { 
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;++i) scanf("%d",&a[i]);
    for(int i = 0;i <= up;++i) r[i] = up,G[i].push_back(0);
    for(int i = 1;i <= n;++i) {
        for(auto v : G[a[i]]) {
            int ma = a[i] ^ v;
            vis[ma] = 1;
            while(r[ma] > a[i]) {
                r[ma]--;
                if(r[ma] != a[i]) G[r[ma]].push_back(ma);
            }
        }
        G[a[i]].clear();
    }    
    int ans = 1;
    for(int i = 1;i <= up;++i) {
        ans += vis[i];
    }
    printf("%d\n",ans);
    for(int i = 0;i <= up;++i) {
        if(i == 0 || vis[i]) printf("%d ",i);
    }
}   
int main() {
    //int ca;scanf("%d",&ca);
    //while(ca--) {
        solve();
    //}
   // system("pause");
    return 0;
}

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来源： https://www.cnblogs.com/zwjzwj/p/15467023.html