【Codeforces】 Codeforces Round 873 (Div. 2) （ABC）（+63!!!）-优快云博客

本文链接：https://blog.youkuaiyun.com/t_mod/article/details/130676708

A Divisible Array

解题思路

$1+2+3+4+...+n=\frac{n(n+1)}{2}$

$2+4+6+...+n=n(n+1)$

很明显第二个式子满足题目条件

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define fornk(i, l, r, k) for (int i = l; i <= r; i += k)
const int N = 2e5 + 5;
ll T;
ll n;
ll a[N];
void solve() {
	cin >> n;
	forn(i, 1, n) cout << 2 * i << ' ';
	cout << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}

B Permutation Swap

解题思路

$\left | a_i-i \right |$ 就是 $a_i$ 离正确位置的距离，题目让我们求使用最大的移动步数移动全部到正确的位置，那么这个最大移动距离的既是所有 $\left | a_i-i \right |$ 的最大公约数

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define fornk(i, l, r, k) for (int i = l; i <= r; i += k)
const int N = 2e5 + 5;
ll T;
ll n;
ll a[N];
ll ans;
void solve() {
	ans = 0;
	cin >> n;
	forn(i, 1, n) cin >> a[i];
	forn(i, 1, n) ans = __gcd( abs(a[i] - i), ans);
	cout << ans << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}

C Counting Orders

解题思路

我们定义数组c为a数组中严格大于 $b_i$ 的个数，那么从从第一个数开始我们有 $c_1$ 个数可以选，接下来是选 $c_2-1$ ，... $c_3-2$ 那么答案既是 $\prod_{i=1}^{n} {c_i-i+1}$ ,注意对 $c_i$ 小于1的情况特判

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define fornk(i, l, r, k) for (int i = l; i <= r; i += k)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define fornk_(i, l, r, k) for (int i = l; i >= r; i -= k)
const int N = 2e5 + 5;
ll T;
ll n;
ll a[N], b[N], c[N];
ll ans;
ll mod = 1e9 + 7;
bool cmp(ll a, ll b) {
	return a > b;
}
void solve() {
	ans = 1;
	cin >> n;
	forn(i, 1, n) cin >> a[i];
	forn(i, 1, n) cin >> b[i];
	sort(a + 1, a + n + 1, cmp);
	sort(b + 1, b + 1 + n);
	ll cnt = n;
	forn(i, 1, n) {
		while (a[cnt] <= b[i]) cnt--;
		c[i] = cnt;
	}
	cnt = 0;
	forn_(i, n, 1) {
		if (c[i] - cnt < 1) ans = 0;
		ans *= (c[i] - cnt++);
		ans %= mod;
	}
	cout << ans << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}