A

#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 turn[][] = {{, }, { -, }, {, }, {, -}};

//priority_queue<int, vector<int>, less<int>> que;

//next_permutation

ll Mod = ;

int num[];

int pre[];

int n;

int number;

bool check()

{

        for (int i = ; i <= n; i++)

        {

                pre[i] = pre[i - ] + pre[i];

                if (pre[i] < )

                {

                        return false;

                }

        }

        return true;

}

int main()

{

        int T;

        cin >> T;

        while (T--)

        {

                cin >> n >> number;

                for (int i = ; i <= number; i++)

                {

                        cin >> num[i];

                }

                for (int i = ; i <= ; i++)

                {

                        for (int j = ; j <= n; j++)

                        {

                                pre[j] = ;

                        }

                        for (int j = ; j <= number; j++)

                        {

                                if (i == )

                                {

                                        pre[num[j]] += ;

                                        pre[num[j] + ] += -;

                                }

                                else

                                {

                                        pre[max(, num[j] - i + )] += ;

                                        pre[num[j] + i] += -;

                                }

                        }

                        if (check())

                        {

                                cout << i << endl;

                                break;

                        }

                }

        }

        return ;

}

B

#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-8;

typedef pair<int, int> pairint;

typedef long long ll;

typedef unsigned long long ull;

//const int maxn = 3e5 + 10;

const int  maxm = ;

const int turn[][] = {{, }, { -, }, {, }, {, -}};

//priority_queue<int, vector<int>, less<int>> que;

//next_permutation

ll mod = 3e7;

int le[];

int re[];

int wait[];

int main()

{

        int t;

        cin >> t;

        while (t--)

        {

                int n;

                scanf("%d", &n);

                for (int i = ; i < n; i++)

                {

                        scanf("%d %d", &le[i], &re[i]);

                }

                int cur = le[];

                for (int i = ; i < n; i++)

                {

                        cur = max(cur, le[i]);

                        if (re[i] >= cur)

                        {

                                wait[i] = cur;

                                cur++;

                        }

                        else

                        {

                                wait[i] = ;

                        }

                }

                for (int i = ; i < n; i++)

                {

                        cout << wait[i];

                        if (i != n - )

                        {

                                cout << " ";

                        }

                }

                cout << endl;

        }

        return ;

}

C

#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 turn[][] = {{, }, { -, }, {, }, {, -}};

//priority_queue<int, vector<int>, less<int>> que;

//next_permutation

ll Mod = ;

int num[];

int pre[];

char f[];

int main()

{

        int n;

        cin >> n;

        for (int i = ; i <= n; i++)

        {

                scanf("%d", num + i);

        }

        scanf("%s", f + );

        for (int i = ; i < n; i++)

        {

                if (f[i] == '')

                {

                        pre[i] = pre[i - ] + ;

                }

                else

                {

                        pre[i] = ;

                }

        }

        for (int i = ; i < n; i++)

        {

                if (pre[i] > pre[i + ])

                {

                        sort(num + i - pre[i] + , num + i + );

                }

        }

        for (int i = ; i < n; i++)

        {

                if (num[i] != i)

                {

                        cout << "NO" << endl;

                        return ;

                }

        }

        cout << "YES" << endl;

        return ;

}

E

把所有点放在一个set里，每次取set中一个顶点，删去，遍历set，删去与此顶点邻接的顶点

因为遍历的过程中有两种结局1.删去某个结点 遍历成功 2.两点之间不存在边 遍历失败

所以遍历的总复杂度为O(n+m) 再加上set的复杂度就是 O((n+m)log)

#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 turn[][] = {{, }, { -, }, {, }, {, -}};

//priority_queue<int, vector<int>, less<int>> que;

//next_permutation

ll Mod = ;

set<int> need;

map<int, bool> mp[];

vector<int> ans;

int main()

{

        int n, m;

        int from, to;

        cin >> n;

        for (int i = ; i <= n; i++)

        {

                need.insert(i);

        }

        cin >> m;

        for (int i = ; i <= m; i++)

        {

                scanf("%d %d", &from, &to);

                mp[from][to] = mp[to][from] = ;

        }

        while (!need.empty())

        {

                int todo = *need.begin();

                need.erase(todo);

                queue<int> que;

                ans.push_back();

                que.push(todo);

                while (!que.empty())

                {

                        queue<int> shan;

                        int cnt = que.front();

                        que.pop();

                        ans.back()++;

                        for (auto i : need)

                        {

                                if (!mp[cnt][i])

                                {

                                        que.push(i);

                                        shan.push(i);

                                }

                        }

                        while (!shan.empty())

                        {

                                need.erase(shan.front());

                                shan.pop();

                        }

                }

        }

        cout << ans.size() << endl;

        sort(ans.begin(), ans.end());

        for (auto i : ans)

        {

                cout << i << " ";

        }

        cout << endl;

        return ;

}

F

如果知道到N的因数(N%i==0)数量级是N^(1/3)的这道题就很好做了 因为当N=1或者N=2时因数数目等于N 而1e6=2^20 每个数最多被修改7次

所以线段树维护一个最大值 一个sum值 当最大值不大于2时不用修改 大于二时递归下去暴力修改

#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 turn[][] = {{, }, { -, }, {, }, {, -}};

//priority_queue<int, vector<int>, less<int>> que;

//next_permutation

ll Mod = ;

ll n, q;

int l, r;

ll dp[];

struct node

{

        ll maxn, sum;

} tree[];

void pushup(int x)

{

        tree[x].sum = tree[x << ].sum + tree[x <<  | ].sum;

        tree[x].maxn = max(tree[x << ].maxn, tree[x <<  | ].maxn);

}

void build(int x, ll value, int root = , int l = , int r = n)

{

        if (l > x || r < x)

        {

                return ;

        }

        if (l == x && r == x)

        {

                tree[root].sum = tree[root].maxn = value;

                return;

        }

        int mid = (l + r) >> ;

        if (x <= mid)

        {

                build(x, value, root << , l, mid);

        }

        else

        {

                build(x, value, root <<  | , mid + , r);

        }

        pushup(root);

}

void update(int xl, int xr, int root = , int l = , int r = n)

{

        if (l > r || l > xr || r < xl)

        {

                return;

        }

        if (xl <= l && xr >= r && tree[root].maxn <= )

        {

                return;

        }

        if (l == r)

        {

                tree[root].sum = tree[root].maxn = dp[tree[root].sum];

                return ;

        }

        int mid = (l + r) >> ;

        if (xl <= mid)

        {

                update(xl, xr, root << , l, mid);

        }

        if (xr > mid)

        {

                update(xl, xr, root <<  | , mid + , r);

        }

        pushup(root);

}

ll getsum(int xl, int xr, int root = , int l = , int r = n)

{

        if (l > r || l > xr || r < xl)

        {

                return ;

        }

        if (xl <= l && xr >= r)

        {

                return tree[root].sum;

        }

        int mid = (l + r) >> ;

        return getsum(xl, xr, root << , l, mid) + getsum(xl, xr, root <<  | , mid + , r);

}

int main()

{

        cin >> n >> q;

        ll cnt;

        for (int i = ; i <= ; i++)

        {

                for (int j = i; j <= ; j += i)

                {

                        dp[j]++;

                }

        }

        for (int i = ; i <= n; i++)

        {

                scanf("%lld", &cnt);

                build(i, cnt);

        }

        for (int i = ; i <= q; i++)

        {

                int now;

                ll value;

                int aim;

                cin >> now;

                if (now == )

                {

                        scanf("%d %d", &l, &r);

                        update(l, r);

                }

                else

                {

                        scanf("%d %d", &l, &r);

                        cout << getsum(l, r) << endl;

                }

        }

        return ;

}

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