Codeforces 939 时区模拟 三分

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 ;
}