Codeforces Round #703 (Div. 2)补题

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前言

一题自闭,掉大分。

题目

A

注意只能后移不能前移即可。

B

之前学长讲过一维推广到二维的情况,赛时也回想到了,但是没想到曼哈顿距离也能推广。算是个结论吧。

一维:奇数取中位数,偶数取两个中位数之间的任何位置都可以。

AC代码:

/*
 * @Author: hesorchen
 * @Date: 2020-11-26 09:12:46
 * @LastEditTime: 2021-02-19 15:31:36
 * @Description: 栽种绝处的花
 */
#include <bits/stdc++.h>
using namespace std;

int x[1010];
int y[1010];

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
            cin >> x[i] >> y[i];
        sort(x + 1, x + 1 + n);
        sort(y + 1, y + 1 + n);
        if (n & 1)
            cout << 1 << endl;
        else
            cout << 1ll * (abs(x[n / 2] - x[n / 2 + 1]) + 1) * (abs(y[n / 2] - y[n / 2 + 1]) + 1) << endl;
    }

    return 0;
}

C2

交互题限制次数。1e5 20次,显然二分。

赛时炸了脑子不够清醒,在那码了删,删了码也不知道码了些啥,比赛一结束理了一下思路就知道正解了。

先查询一次[1,n],得到position

  • position==1,那么最大值肯定在[2,n],对这个区间进行二分,每次查询[1,mid],如果结果不为1,那么说明最大值在[mid+1,n],否则最大值在[2,mid]
  • position==n,那么最大值肯定在[1,n-1],对这个区间进行二分,每次查询[mid,n],如果结果不为n,那么说明最大值在[1,mid-1],否则最大值在[mid,n-1]
  • position在[2,n-1],我们再查询一次[1,position],如果结果还是position,那么最大值在区间[1,position],我们把position看做n,按照情况二处理即可。否则把position看做1,按照情况一处理。

AC代码:

/*
 * @Author: hesorchen
 * @Date: 2020-11-26 09:12:46
 * @LastEditTime: 2021-02-19 15:39:29
 * @Description: 栽种绝处的花
 */
#include <bits/stdc++.h>
using namespace std;

int main()
{
    long long n;
    cin >> n;
    int pre = n, ans, flag = 1, res;
    cout << "? " << 1 << ' ' << n << endl;
    cout.flush();
    cin >> pre;
    if (n == 2)
    {
        int l = 1, r = 2;
        cout << "? " << l << ' ' << r << endl;
        cout.flush();
        cin >> pre;
        if (pre == l)
            return cout << "! " << r << endl, 0;
        else
            return cout << "! " << l << endl, 0;
    }
    if (pre == 1) //情况1
    {
        int l = 2, r = n;
        ans = n;
        while (l <= r)
        {
            int mid = l + r >> 1;
            cout << "? " << pre << ' ' << mid << endl;
            cout.flush();
            int temp;
            cin >> temp;
            if (temp != pre)
                l = mid + 1;
            else
            {
                r = mid - 1;
                ans = mid;
            }
        }
    }
    else if (pre == n) //情况2
    {
        int l = 1, r = n - 1;
        ans = 1;
        while (l <= r)
        {
            int mid = l + r >> 1;
            cout << "? " << mid << ' ' << pre << endl;
            cout.flush();
            int temp;
            cin >> temp;
            if (temp != pre)
                r = mid - 1;
            else
            {
                l = mid + 1;
                ans = mid;
            }
        }
    }
    else //情况3
    {
        int l = 1, r = n;
        int re1, re2;
        cout << "? " << 1 << ' ' << pre << endl;
        cout.flush();
        cin >> re1;

        if (re1 == pre) //类比情况2
        {
            int l = 1, r = pre - 1;
            ans = 1;
            while (l <= r)
            {
                int mid = l + r >> 1;
                cout << "? " << mid << ' ' << pre << endl;
                cout.flush();
                int temp;
                cin >> temp;
                if (temp != pre)
                    r = mid - 1;
                else
                {
                    l = mid + 1;
                    ans = mid;
                }
            }
        }
        else //类比情况1
        {
            int l = pre + 1, r = n;
            ans = r;
            while (l <= r)
            {
                int mid = l + r >> 1;
                cout << "? " << pre << ' ' << mid << endl;
                cout.flush();
                int temp;
                cin >> temp;
                if (temp != pre)
                    l = mid + 1;
                else
                {
                    r = mid - 1;
                    ans = mid;
                }
            }
        }
    }
    cout << "! " << ans << endl;
    return 0;
}

打比赛时头脑还是不够清醒,得改掉这个坏习惯。