Codeforces 934 最长不递减子序列 多项式系数推导

A

B

C

给你一个长度为N的01串 你可以翻转一次任意【L，R】的区间

问你最长的不递减序列为多少长

处理出1的前缀和 和2的后缀和

然后N^2 DP 处理出 【L,R】区间的最长不递增序列

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
const int MAXN = 2005;
int a[2005];
int pre[2005];
int suf[2005];
int dp[2005][2005][3];
int ans = 0;
int main()
{
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                cin >> a[i];
        }
        for (int i = 1; i <= n; i++)
        {
                pre[i] = pre[i - 1] + (a[i] == 1);
        }
        for (int i = n; i >= 1; i--)
        {
                suf[i] = suf[i + 1] + (a[i] == 2);
        }
        for (int i = 1; i <= n; i++)
        {
                ans = max(pre[i] + suf[i], ans);
        }
        //cout<<ans<<endl;
        for (int i = 1; i <= n; i++)
        {
                for (int j = i; j <= n; j++)
                {
                        dp[i][j][2] = dp[i][j - 1][2] + (a[j] == 2);
                        dp[i][j][1] = max(dp[i][j - 1][2], dp[i][j - 1][1]) + (a[j] == 1);
                        ans = max(ans, max(dp[i][j][2], dp[i][j][1]) + pre[i - 1] + suf[j + 1]);
                }
        }
        cout << ans << endl;
}

D

找规律 

考虑构造q(x).先让常数项小于k,也就是让相乘后的常数项+p小于k.

q(x)的常数项就是p / (-k).这样会产生一次项，那么继续消一次项.直到最后的p = 0.

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = 1000000007;
int cnt;
int ans[1000005];
int main()
{
        ll p;
        ll k;
        cin >> p >> k;
        ll flag = 0;
        while (p != 0)
        {
                ans[++cnt] = p % k;
                p = p / k * (-1);
                if (flag)
                {
                        p += flag;
                        flag = 0;
                }
                if (p < 0)
                {
                        flag += (k - p) / k;
                        p += flag * k;;
                }
        }
        cout << cnt << endl;
        for (int i = 1; i <= cnt; i++)
        {
                cout << ans[i] << " ";
        }
        cout << endl;
        return 0;
}