Codeforces Round 1006 (Div. 3)(部分题解)

A. New World, New Me, New Array

思路：这题挺简单的，在对k取绝对值后会更简单一点。

然后注意上取整。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int n, k, p;
    cin >> n >> k >> p;
    k = abs(k);
    if (n*p < k) cout << -1 << endl;
    else {
        cout << (k+p-1)/p << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

B. Having Been a Treasurer in the Past, I Help Goblins Deceive

思路：统计‘-’和‘_’的数量，‘-’会分在‘_’的两边，保证左右乘积最大，即尽可能平均分配。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int n;
    string s;
    cin >> n;
    cin >> s;
    int n1 = count(s.begin(), s.end(), '-');
    int n2 = n-n1;
    if (n1 < 2 || n2 < 1) cout << 0 << endl;
    else {
        int cnt = n1/2;
        cout << cnt*(n1-cnt)*n2 << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

C. Creating Keys for StORages Has Become My Main Skill

思路：我看了半天才看懂MEX(S)的意思，这里补充介绍一下，就是它的值为0到后第一个没在S集合里面的值，因此它的最大值不会超过n-1，所以我们要让它最大，就要从0开始枚举。这里代码(i&x)==i表示i的二进制数都包含在x中，一旦枚举到不满足，MEX值定了，所以我们可以为任何(i&x)==i的i值，显然0必定符合且简单，所以不满足就输出0，当枚举到最后一个数时，我们还要考虑前面所有的OR是否等于x，这里我们考虑到如果加上次i=n-1满足也可以输出该i，不满足我们就要输出一个能让其满足的数，即(num^x)可以帮缺少的二进制1都补上。（二进制）

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int n, x;
    cin >> n >> x;
    int num = 0;
    for (int i = 0; i<n; i++){
        if (i == n-1){
            if ((i|num) == x){
                cout << i << endl;
            }
            else cout << (num^x) << endl;
            return ;
        }
        if ((i&x)==i){
            cout << i << " ";
            num |= i;
        }
        else{
            cout << 0 << " ";
        }
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

D. For Wizards, the Exam Is Easy, but I Couldn't Handle It

思路：这道题由于n的范围比较小，可以直接暴力做出来，对于一个位置，往后枚举比较。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i<n; i++) cin >> a[i];
    int sum = 0;
    for (int i = 0; i<n; i++){
        for (int j = i+1; j<n; j++){
            sum += (a[i] > a[j]?1:0);
        }
    }
    int l = 0, r = 0;
    int ans = sum;
    for (int i = 0; i<n; i++){
        int cur = sum;
        for (int j = i+1; j<n; j++){
            cur -= (a[i] > a[j]?1:0);
            cur += (a[i] < a[j]?1:0);
            if (cur < ans){
                ans = cur;
                l = i;
                r = j;
            }
        }
    }
    cout << l+1 << " " << r+1 << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

E. Do You Love Your Hero and His Two-Hit Multi-Target Attacks?

思路：首先看到有公式之类要想着化简或者转化，这里对公式化简为

所以只要x轴相等或者y轴相等就可以为一对，我们可以指定一个轴，然后来满足此条件。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    int k;
    cin >> k;
    vector<array<int, 2>> ans;
    int x = 0, y = 0;
    while (k){
        x++;
        int s = 1;
        while (s*(s+1)/2 <= k) s++;
        k -= s*(s-1)/2;
        for (int i = 1; i<=s; i++){
            y++;
            ans.push_back({x, y});
        }
    }
    cout << ans.size() << endl;
    for (auto [x, y]:ans){
        cout << x << " " << y << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}