Codeforces 948 数论推导融雪前缀和二分check 01字典树带删除

最新推荐文章于 2024-10-07 02:38:27 发布

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最新推荐文章于 2024-10-07 02:38:27 发布

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原文链接：http://www.cnblogs.com/Aragaki/p/8554259.html

全部空的放狗

先O（NLOGNLOGN）处理出一个合数质因数中最大的质数是多少

因为p1 x1 x2的关系是 x2是p在x1之上的最小倍数所以x1的范围是[x2-p+1,x2-1]要使最后答案尽可能小要包含尽可能多的选择

p0 x0 x1关系同上

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#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  maxn = 100005;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
const int turn2[8][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}, {1, -1}, { -1, -1}, {1, 1}, { -1, 1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 3e7;
int prime[1000005];
//priority_queue<int, vector<int>, less<int>> que;
void init()
{
        for (int i = 2; i <= 1000000; i++)
        {
                if (prime[i])
                {
                        continue;
                }
                for (int j = i * 2; j <= 1000000; j += i)
                {
                        prime[j] = i;
                }
        }
}
int main()
{
        int n;
        init();
        cin >> n;
        int anser = INT_MAX;
        int now = prime[n];
        //cout << prime[n] << endl;
        int x1 = n - now + 1;
        for (int i = x1; i <= n; i++)
        {
                if (!prime[i])
                {
                        continue;
                }
                anser = min(anser, i - prime[i] + 1);
                //cout << i - prime[i] + 1 << endl;
        }
        cout << anser << endl;
}

View Code

前缀和题

作温度的前缀和

二分出第i块雪在第j天融化

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#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
ll mod = 3e7;
const int maxn=1e5+5;
ll v[maxn];
ll t[maxn];
ll pre[maxn];
ll ans[maxn];
ll add[maxn];
int main()
{
       ll n;
       cin >> n;
       for(int i=1;i<=n;i++)
       {
               scanf("%lld",v+i);
       }
       for(int i=1;i<=n;i++)
       {
               scanf("%lld",t+i);
               pre[i]=pre[i-1]+t[i];
       }
       pre[n+1]=2e18+1;
       for(int i=1;i<=n;i++)
       {
               ll now=v[i]+pre[i-1];
               int aim=upper_bound(pre,pre+n+1,now)-pre;
               //cout<<aim<<endl;
               if(aim==n+1)
                continue;
                add[aim]+=t[aim]-(pre[aim]-now);
                //cout<<t[aim]-(pre[aim]-now)<<endl;
                ans[aim]--;
       }
       for(int i=1;i<=n;i++)
       {
               ans[i]+=ans[i-1];
       }
       for(int i=1;i<=n;i++)
       {
               ll anser=(ans[i]+i)*t[i]+add[i];
               cout<<anser<<" ";
       }
       cout<<endl;

}

View Code

D.01字典树带删除路径(用数组维护)

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#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
ll mod = 3e7;
const int maxn = 3e5 + 5;
int n, m;
int ch[32 * maxn][2];
int sum[32 * maxn];
int a[maxn];
int b[maxn];
int node_cnt;
inline void read(int &jqk)
{
        jqk = 0;
        char c = 0;
        int p = 1;
        while (c < '0' || c > '9')
        {
                if (c == '-')
                {
                        p = -1;
                }
                c = getchar();
        }
        while (c >= '0' && c <= '9')
        {
                jqk = (jqk << 3) + (jqk << 1) + c - '0';
                c = getchar();
        }
        jqk *= p;
}
void Insert(int x)
{
        int cur = 0;
        for (int i = 30; i >= 0; i--)
        {
                int idx = (x >> i) & 1;
                if (!ch[cur][idx])
                {
                        //ch[node_cnt][1] = ch[node_cnt][0] = 0;
                        ch[cur][idx] = ++node_cnt;
                }
                cur = ch[cur][idx];
                sum[cur]++;
        }
}
int getans(int x)
{
        int anser = 0;
        int cur = 0;
        for (int i = 30; i >= 0; i--)
        {
                int idx = (x >> i) & 1;
                if (!ch[cur][idx] || !sum[ch[cur][idx]])
                {
                        idx ^= 1;
                        anser += (1 << i);
                }
                cur = ch[cur][idx];
                sum[cur]--;
        }
        return anser;
}
int main()
{
        int n;
        read(n);
        for (int i = 1; i <= n; i++)
        {
                read(a[i]);
        }
        for (int i = 1; i <= n; i++)
        {
                read(b[i]);
                Insert(b[i]);
        }
        for (int i = 1; i <= n; i++)
        {
                cout << getans(a[i]) << " ";
        }
        cout << endl;
}