Codeforces Round #618 (Div. 2)

题库链接

A. Non-zero

一个数组,每次操作可以给某个数加1,让这个数组的积和和不为0的最小操作数

显然如果有0的话,必须操作一次,最后如果和还是为0的话,再操作一次

#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))

#define pii pair<int,int>

#define int long long

#define gcd __gcd

const int inf = 0x3f3f3f3f;

const int maxn = 201110;

const int M = 1e9+7;

int n,m;

int a[maxn];

signed main()

{

#ifdef ONLINE_JUDGE

#else

    freopen("data.in", "r", stdin);

#endif

    int T;

    cin>>T;

    while(T--)

    {

        cin>>n;

        int sum = 0;

        int ans = 0;

        for(int i = 1; i <= n; i++)

        {

            cin>>a[i];

            if(a[i] == 0)

            {

                ans++;

                a[i] = 1;

            }

            sum += a[i];

        }

        if(sum == 0) ans++;

        cout<<ans<<endl;

    }

    return 0;

}

B. Assigning to Classes

有一个数组有2*n个元素,要分成两个数组,每个数组必须是奇数个元素,求两个数组的中位数之差的最小值

假设左边的中位数会小于右边的中位数,显然,左边越大越好,右边越小越好,大能大到什么程度呢?最大就是原数组的中位数

所以答案就是原数组的中间两个数的差值

#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))

#define pii pair<int,int>

#define int long long

#define gcd __gcd

const int inf = 0x3f3f3f3f;

const int maxn = 201110;

const int M = 1e9+7;

int n,m;

int a[maxn];

signed main()

{

#ifdef ONLINE_JUDGE

#else

    freopen("data.in", "r", stdin);

#endif

    int T;

    cin>>T;

    while(T--)

    {

        cin>>n;

        for(int i = 1; i <= 2*n; i++)

        {

            cin>>a[i];

        }

        sort(a+1,a+1+2*n);

        cout<<abs(a[n]-a[n+1])<<endl;

    }

    return 0;

}

C. Anu Has a Function

$f(x, y) = (x | y) - y$

要求重新排列数组,使得$f(f(\dots f(f(a_1, a_2), a_3), \dots a_{n-1}), a_n)$最大

我们先看f函数,x或y减去y,把它拆分成二进制,对于某一位,如果x,y都是1,f(x, y)之后x的这一位会被减去

如果x是1,y是0,或者x是0,y是1,那么x的这一位不变

所以我们只需要求出某一位只有x是1,其余数组元素都是0的最高位,让它在最前面就好了,其它元素随便排列

#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))

#define pii pair<int,int>

#define int long long

#define gcd __gcd

const int inf = 0x3f3f3f3f;

const int maxn = 201110;

const int M = 1e9+7;

int n,m;

int a[maxn];

int vis[33];

int b[33];

signed main()

{

#ifdef ONLINE_JUDGE

#else

    freopen("data.in", "r", stdin);

#endif

    cin>>n;

    for(int i = 1; i <= n; i++)

    {

        cin>>a[i];

        for(int j = 30; j >= 0; j--)

        {

            if((1<<j)&a[i])

            {

                vis[j]++;

                b[j] = i;

            }

        }

    }

    int ans = 1;

    for(int i = 0; i <= 30; i++)

    {

        if(vis[i] == 1) ans = b[i];

    }

    cout<<a[ans]<<' ';

    for(int i = 1; i <= n; i++)

    {

        if(i != ans) cout<<a[i]<<' ';

    }

    return 0;

}

D. Aerodynamic

就是问你是否中心对称

#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))

#define pii pair<int,int>

#define int long long

#define gcd __gcd

const int inf = 0x3f3f3f3f;

const int maxn = 201110;

const int M = 1e9+7;

int n,m;

vector<pii> v;

signed main()

{

#ifdef ONLINE_JUDGE

#else

    freopen("data.in", "r", stdin);

#endif

    cin>>n;

    v.push_back({0,0});

    for(int i = 1,x,y; i <= n; i++)

    {

        cin>>x>>y;

        v.push_back({x,y});

    }

    if(n%2) {puts("NO");return 0;}

    int t = n/2;

    double midx = (v[1].first + v[t+1].first)*1.0/2;

    double midy = (v[1].second + v[t+1].second)*1.0/2;

    for(int i = 2; i <= t; i++)

    {

        double tmpx = (v[i].first + v[t+i].first)*1.0/2;

        double tmpy = (v[i].second + v[t+i].second)*1.0/2;

        if(midx != tmpx || midy != tmpy)

        {

            puts("NO");return 0;

        }

    }

    puts("YES");

    return 0;

}

E. Water Balance

有n个水桶,每次可以选择一些区间,使得区间内的水桶的水量等于区间水量之和除以区间长度

问你可以得到的水量的字典序的最小是什么

单调栈可以做,凸包也可以做,凸包我不会,说一下单调栈的做法

#include <bits/stdc++.h>

using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))

#define pii pair<int,int>

#define int long long

#define gcd __gcd

const int inf = 0x3f3f3f3f;

const int maxn = 1001110;

const int M = 1e9+7;

int n,m;

int a[maxn];

int sa[maxn];

int la[maxn];

signed main()

{

#ifdef ONLINE_JUDGE

#else

    freopen("data.in", "r", stdin);

#endif

    cin>>n;

    for(int i = 1; i <= n; i++)

    {

        cin>>a[i];

    }

    int top = 0;

    for(int i = 1; i <= n; i++)

    {

        int tsa = a[i],tla = 1;

        for(;top && sa[top]*tla >= tsa*la[top]; top--)

        {

            tsa += sa[top];

            tla += la[top];

        }

        ++top;

        sa[top] = tsa;

        la[top] = tla;

    }

    for(int i = 1,cur = 1; i <= top; i++,cur += la[top])

    {

        for(int j = cur;j < cur + la[i]; j++)

        {

            printf("%.9f\n",sa[i]*1.0/la[i]);

        }

    }

    return 0;

}