Educational Codeforces Round 11

Educational Codeforces Round 11

https://codeforces.com/contest/660
4/6：ABCD

A. Co-prime Array

\(1\) 与任何数的 \(gcd\) 都为 \(1\)，直接在不符合条件的两点间塞就可以了

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

using namespace std;
const int N = 1005;
int a[N], n, ans;
bool vis[N];

void solve () {
    cin >> n;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    for (int i = 2; i <= n; i++) {
        if (__gcd (a[i-1], a[i]) != 1)  vis[i-1] = true, ans ++;
    }
    cout << ans << endl;
    for (int i = 1; i <= n; i++) {
        cout << a[i] << ' ';
        if (vis[i])     cout << "1 ";
    }
}

signed main () {
    solve ();
}

B. Seating On Bus

仔细读题，注意细节。

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

using namespace std;

void solve () {
    int n, m;
    cin >> n >> m;
    if (m <= 2 * n) {
        for (int i = 1; i <= m; i++)    cout << i << ' ';
        return ;
    }
    for (int i = 1; i <= 2 * n; i++) {
        if (i + 2 * n <= m)     cout << i + 2 * n << ' ';
        cout << i << ' ';
    }
}

signed main () {
    solve ();
}

C. Hard Process

此题有很多做法。
这里用双指针主要是考虑从暴力转移，区间 \([l,r]\) 范围内 \(0\) 的个数，\(\leq k\)，不难发现这是一个区间和问题，故可以用双指针来维护，和这题 D - Enough Array 很类似

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

using namespace std;
typedef pair<int, int> pii;
const int N = 3e5 + 5;
int n, k, a[N];

void solve () {
    cin >> n >> k;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    int l = 1, r = 1, sum = 0, ans = 0, ansl, ansr;
    while (r <= n) {
        sum += !a[r]; //0的个数
        if (sum > k)        sum -= !a[l ++];
        if (r - l + 1 > ans) {
            ans = r - l + 1;
            ansl = l, ansr = r;
        }
        r ++;
    }
    cout << ans << endl;
    for (int i = ansl; i <= ansr; i++)  a[i] = 1;
    for (int i = 1; i <= n; i++)    cout << a[i] << ' ';
}

signed main () {
    solve ();
}

D. Number of Parallelograms

利用平行四边形对角线互相平分的性质，记录两点之间的中点，即可最简化代码。

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

using namespace std;
typedef pair<int, int> pii;
const int N = 2005;
pii p[N];
int n, ans;

void solve () {
    cin >> n;
    for (int i = 1; i <= n; i++)    cin >> p[i].first >> p[i].second;
    map<pii, int> mp;
    for (int i = 1; i < n; i++) {
        for (int j = i + 1; j <= n; j++) {
            mp[{p[i].first + p[j].first, p[i].second + p[j].second}] ++;
        }
    }
    for (auto i : mp) {
        ans += (i.second - 1) * i.second / 2;
    }
    
    cout << ans << endl;
}

signed main () {
    solve ();
}

E. Different Subsets For All Tuples

组合数学推式子。
结论：

\[\sum_{i=0}^{n-1}m^{n-i}(2m-1)^j \]

推导：https://www.cnblogs.com/chenxiaoran666/p/CF660E.html

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

using namespace std;
typedef pair<int, int> pii;
const int p = 1e9 + 7;
int n, m;

int qmi(int a, int k) {
    if (k == 0 || a == 1)    return 1;
    int ans = 1;
    while (k) {
        if (k & 1)    ans = ans * a % p;
        k >>= 1;
        a = a * a % p;
    }
    return ans;
}

void solve() {
    cin >> n >> m;
    int ans1 = qmi (m, n), ans2 = 1, dx1 = qmi (m, p - 2), dx2 = 2 * m - 1;
    int ans = ans1 * ans2;
    for (int i = 0; i < n; i++) {
        ans = (ans + ans1 * ans2) % p;
        ans1 = (ans1 * dx1) % p, ans2 = (ans2 * dx2) % p; //递推
    }
    cout << ans;
}

signed main() {
    solve();
}

//sum: m^{n-i}(2m-1)^i, i从0到n-1

F. Bear and Bowling 4

斜率优化