AcWing2022春季每日一题-2-第四周

AcWing2022春季每日一题，第二周题解

AcWing 1812. 方形牧场—原题链接

题目标签：枚举

思路：
暴力枚举即可

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int x1, y1, x2, y2;
    x1 = y1 = 15, x2 = y2 = 0;
    int a1, b1, c1, d1, a2, b2, c2, d2;
    cin >> a1 >> b1 >> c1 >> d1 >> a2 >> b2 >> c2 >> d2;
    x1 = min(x1, min(a1, min(c1, min(a2, c2))));
    y1 = min(y1, min(b1, min(d1, min(b2, d2))));
    x2 = max(x2, max(a1, max(c1, max(a2, c2))));
    y2 = max(y1, max(b1, max(d1, max(b2, d2))));
    
    int n = max(abs(x2 - x1), abs(y2 - y1));
    cout << n * n << endl;
    return 0;
}

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AcWing 1800. 不做最后一个！—原题链接

题目标签：

思路：
n不大于100，因此直接暴力，非常滴暴力且不聪明

#include <bits/stdc++.h>
using namespace std;daaaaaaa
typedef long long LL;
typedef pair<int, int> PII;
int n;

void solve();

int main()
{
    cin >> n;
    unordered_map<string, int> mp = {{"Bessie", 0},{"Elsie", 0},{"Daisy", 0},{"Gertie", 0},{"Annabelle", 0},{"Maggie", 0},{"Henrietta", 0}};
    while(n--)
    {
        string s;
        int x;
        cin >> s >> x;
        mp[s] += x;
    }

    vector<pair<string, int>> v1;
    int minn = 0x3f3f3f3f;
    for(auto t : mp)
    {
        minn = min(minn, t.second);
    }
    for(auto t : mp)
    {
        if(t.second != minn) v1.push_back(t);
    }
    if(v1.size() == 0)
    {
        puts("Tie");
        return 0;
    }
    vector<pair<string, int>> v2;
    minn = 0x3f3f3f3f;
    for(int i=0; i<v1.size(); i++)
    {
        minn = min(minn, v1[i].second);
    }
    for(int i=0; i<v1.size(); i++)
    {
        auto t = v1[i];
        if(t.second == minn) v2.push_back(t);
    }
    if(v2.size() == 1) cout << v2[0].first << endl;
    else puts("Tie");
    // system("pause");
    return 0;
}

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AcWing 1788. 牛为什么过马路—原题链接

题目标签：模拟

思路：
直接模拟

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int N = 11;
const int INF = 0x3f3f3f3f;
int n, cnt;
int p[N];

void solve();

int main()
{
    memset(p, -1, sizeof p);
    cin >> n;
    while(n--)
    {
        int a, b;
        cin >> a >> b;
        if(p[a] == -1)
        {
            p[a] = b;
            continue;
        }
        if(p[a] != b)
        {
            p[a] = b;
            cnt++;
        }
    }
    cout << cnt << endl;
    // system("pause");
    return 0;
}

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AcWing 1775. 丢失的牛—原题链接

题目标签：模拟

思路：直接模拟就行

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

int x, y;
int cnt;
int dist, now, last;

void solve();

int main()
{
    cin >> x >> y;

    int dist = 0, step = 1, last = x;
    while(true)
    {
        int next = x + step;
        if((y >= last && y <= next) || (y >= next && y <= last))
        {
            dist += abs(last - y);
            break;
        }
        dist += abs(last - next);
        step *= (-2);
        last = next;
    }
    cout << dist << endl;
    // system("pause");
    return 0;
}

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AcWing 1866. 围栏刷漆—原题链接

题目标签：集合

思路：
虽然可以暴力模拟，但是集合才应该是本题的考点，两个线段集合在确定头部关系的情况下只有包含，重合，不重合三种情况，分类讨论即可

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

int a, b, c, d;

void solve();

int main()
{
    cin >> a >> b >> c >> d;
    if(a > c)
    {
        swap(a, c);
        swap(b, d);
    }
    if(c > d)
    {
        cout << b + d - a - c << endl;
    }
    else if(d < b)
    {
        cout << b - a << endl;
    }
    else
    {
        cout << d - a << endl;
    }
    // system("pause");
    return 0;
}

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AcWing 1854. 晋升计数—原题链接

题目标签：模拟

思路：
模拟 × 阅读理解 √， 并且使用n是为了兼容比赛组别不一定为4的情况

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n;
int a[N], b[N], res[N];

void solve();

int main()
{
    n = 4;
    for(int i=1; i<=n; i++) cin >> a[i] >> b[i];

    res[n] = b[n] - a[n];
    for(int i=n-1; i>=2; i--)
    {
        res[i] = res[i+1] + b[i] - a[i];
    }
    for(int i=2; i<=n; i++) cout << res[i] << endl;
    return 0;
    // system("pause");
    return 0;
}

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