AtCoder Beginner Contest 128

AtCoder Beginner Contest 128

https://atcoder.jp/contests/abc128

A - Apple Pie

#include <bits/stdc++.h>

using namespace std;

int main () {
    int a, b;
    cin >> a >> b;
    cout << (a * 3 + b) / 2;
}

B - Guidebook

#include <bits/stdc++.h>

using namespace std;

struct Node {
    string name;
    int id, num;
    bool operator < (const Node &t) const {
        if (name != t.name)     return name < t.name;
        return num > t.num;
    }
};

int main () {
    vector <Node> v;
    int n;  cin >> n;
    for (int i = 1; i <= n; i++) {
        string s;
        int num;
        cin >> s >> num;
        v.push_back ({s, i, num});
    }
    sort (v.begin (), v.end ());
    //for (auto i : v)    cout << i.name << ' ' << i.num << endl;
    for (auto i : v)    cout << i.id << endl;
}

C - Switches

#include <bits/stdc++.h>

using namespace std;
const int N = 15;
int p[N], a[N][N], b[N], ans;
int n, m;

void dfs (int x) {
    if (x > n) {
        for (int i = 1; i <= m; i++) {
            int sum = 0;
            for (int j = 1; j <= a[i][0]; j++) {
                if (b[a[i][j]])     sum ++;
            }
            if (sum % 2 != p[i])    return ;
        }
        ans ++;
        return ;
    }
    b[x] = 0, dfs (x + 1);
    b[x] = 1, dfs (x + 1);
}

int main () {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        cin >> a[i][0];
        for (int j = 1; j <= a[i][0]; j++) {
            cin >> a[i][j];
        }
    }
    for (int i = 1; i <= m; i++)    cin >> p[i];
    dfs (1);
    cout << ans;
}

//写个暴力试试

D - equeue

贪心

#include <bits/stdc++.h>

using namespace std;
const int N = 55, M = 105;
int a[N];// f[N][N][M]; //f[i][j][k]: 左边选i个,右边选j个,总代价为k
int n, m, ans, sum1[N], sum2[N];

int main () {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    for (int i = 1; i <= n; i++)    sum1[i] = sum1[i-1] + a[i];
    for (int i = n; i >= 1; i--)    sum2[i] = sum2[i+1] + a[i];
    vector <int> v;
    for (int i = 0; i <= min(n, m); i++) {
        v.clear ();
        for (int j = 1; j <= i; j++)    v.push_back (a[j]);
        for (int j = 0; i + j <= min (n, m); j++) {
            if (j && i != n - j + 1)    v.push_back (a[n-j+1]);
            sort (v.begin (), v.end ());
            //for (auto kk : v)   cout << kk << ' ';cout << endl; 
            int dx = sum1[i] + sum2[n-j+1];
            if (i == n - j + 1)     dx -= a[i];
            for (int k = 0; k < min (m - i - j, (int)v.size ()); k++) {
                if (v[k] >= 0)      break;
                dx -= v[k];
            }
            ans = max (ans, dx);
        }
    }
    cout << ans << endl;
}

E - Roadwork

数据结构题。

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 5;
int n, m, ans[N];
priority_queue<int, vector<int>, greater<int>> add, del; //最近最小坐标

struct Node {
    int op; //0:add, 1:query, 2:del
    int tt, x;
    bool operator<(const Node &t) const {
        if (tt != t.tt)   return tt < t.tt; //先找时间小的
        return op < t.op;
    }
};

int main () {
    vector<Node> v;
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        int st, ed, x;
        cin >> st >> ed >> x;
        st -= x, ed -= (x + 1);
        v.push_back ({0, st, x}), v.push_back ({2, ed, x});
    }
    for (int i = 0; i < m; i++) {
        int tt; cin >> tt;
        v.push_back ({1, tt, i});
    }
    sort (v.begin (), v.end ());

    for (auto i : v) {
        int op = i.op, tt = i.tt, x = i.x;
        if (op == 0)    add.push (x);
        else if (op == 2)   del.push (x);
        else {
            while (!add.empty () && !del.empty () && add.top() == del.top()) {
                add.pop (), del.pop ();
            }
            if (add.empty ())   ans[x] = -1;
            else    ans[x] = add.top ();
        }
    }
    
    for (int i = 0; i < m; i++)     cout << ans[i] << endl;
}

//[s-x,t-x-1]

F - Frog Jump

题意

\(2\) 次为一组，第一次往右移 \(A\) 步，第二次往左移 \(B\) 步，最终到达 \(n−1\)，每个点最多被经过一次，求经过的点的最大和。

分析

固定步长 \(x=A-B\)
往右走：\(0, x, 2x,...\)
往左走：\(n-1, n-1-x, n-1-2x, ...\)
故枚举步长，计算差值，排除越界 + 重复经过。

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 1e5 + 5;
int n, a[N], ans;

signed main () {
    cin >> n;
    for (int i = 0; i < n; i++)     cin >> a[i];
    for (int x = 1; x < n; x++) { //枚举步长
        int sum = 0;
        for (int r = x; r < n; r += x) { //模拟奇数步往右
            int l = n - 1 - r; //模拟偶数步往左
            if (l <= x || (r >= l && l % x == 0))     break; //往左越过起点出界 or 奇偶重复经过
            sum += a[l] + a[r], ans = max (ans, sum);
        }
    }
    cout << ans << endl;
}

//x = A - B
//奇: 0, x, 2x, ... (往右)
//偶：n-1, n-1-x, n-1-2x, ... (往左)