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-8;

typedef pair<int, int> pairint;

typedef long long ll;

typedef unsigned long long ull;

const int maxn = ;

const int  maxm = ;

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

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

//next_permutation

int num[];

int main()

{

    int n;

    cin >> n;

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

    {

        cin >> num[i];

    }

    int flag=;

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

    {

        int time=;

        int aim=num[i];

        while(time--)

        {

            aim=num[aim];

        }

        if(aim==i)

        {

            cout<<"YES"<<endl;

            return ;

        }

    }

    cout<<"NO"<<endl;

    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-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 main()

{

        ll aim=,anser;

        ll remain=1e18+;

        ll n,k;

        cin >> n >> k;

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

        {

                ll now;

                cin >> now;

                if((n-1LL*n/now*now)<remain)

                {

                        aim=i;

                        anser=n/now;

                        remain=n-1LL*n/now*now;

                }

        }

        cout<<aim<<" "<<anser<<endl;

        return ;

}

C

不能用前缀和向后推来判断 应该要向前推

#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 num[];

ll pre[];

ll ans;

ll now;

int main()

{

        int n;

        cin >> n;

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

        {

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

                num[i + n] = num[i];

        }

        ll L, R;

        cin >> L >> R;

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

        {

                ans += num[i];

        }

        now = ans;

        int aim = ;

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

        {

                now = now - num[ + n + R - i] + num[ + n + L - i];

                if (now > ans)

                {

                        aim = i;

                        ans = now;

                }

        }

        cout << aim << endl;

        return ;

}

D

转化题意为求联通块 可以并查集也可以直接DFS 答案为每个联通块内点数-1的和

#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 = ;

string a, b;

int gra[][];

int visit[];

int belong[];

int block[];

void dfs(int x, int aim)

{

        visit[x] = ;

        belong[x] = aim;

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

        {

                if (gra[x][i] && !visit[i])

                {

                        block[aim]++;

                        dfs(i, aim);

                }

        }

}

int main()

{

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

        {

                belong[i] = ;

        }

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

        {

                block[i] = visit[i] = ;

        }

        int n;

        cin >> n;

        cin >> a >> b;

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

        {

                if (a[i] != b[i])

                {

                        visit[a[i] - 'a'] = visit[b[i] - 'a'] = ;

                        gra[a[i] - 'a'][b[i] - 'a'] = gra[b[i] - 'a'][a[i] - 'a'] = ;

                }

        }

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

        {

                if (visit[i])

                {

                        continue;

                }

                dfs(i, i);

        }

        int ans = ;

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

        {

                if (block[i] > )

                {

                        ans += block[i] - ;

                }

        }

        cout << ans << endl;

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

        {

                if (belong[i] !=  && i != belong[i])

                {

                        cout << (char)('a' + i) << " " << (char)('a' + belong[i]) << endl;

                }

        }

        return ;

}

E

裸的一个三分 分析题意可以得到 答案为 ans=maxn-(maxn+pre[k])/(k+1) 为凸函数

也可以记录前一个query的答案然后向后推进

因为随着最大值越来越大maxn-(maxn+pre[k])/(k+1)的值会越来越大 而你要求加进去的前缀里的数只要小于(maxn+pre[k])/(k+1)这个平均值就可以了

#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 num[];

ll pre[];

int pop;

int l, r;

double f(int x)

{

        return ((double)(num[pop]) - (((double)(num[pop]) + (double)(pre[x])) / (double)(x + )));

}

double SF()

{

        while (l < r - )

        {

                int mid = (l + r) >> ;

                int mmid = (mid + r) >> ;

                if (f(mid) > f(mmid))

                {

                        r = mmid;

                }

                else

                {

                        l = mid;

                }

        }

        return max(f(l + ), max(f(l), f(r)));

}

int main()

{

        int q;

        int ch;

        ll number;

        cin >> q;

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

        {

                scanf("%d", &ch);

                if (ch == )

                {

                        scanf("%lld", &number);

                        num[++pop] = number;

                        pre[pop] = pre[pop - ] + number;

                }

                else

                {

                        if (pop == )

                        {

                                cout << "0.00000000" << endl;

                                continue;

                        }

                        if (pop == )

                        {

                                printf("%.9f\n", (double)(num[]) - ((double)num[] + num[]) / 2.0);

                                continue;

                        }

                        l = , r = pop - ;

                        printf("%.9f\n", SF());

                }

        }

        return ;

}

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