AT_arc169_d [ARC169D] Add to Make a Permutation 题解

将取膜去掉，描述满足条件的序列 $B$：

• $\forall i,B_i\ge A_i$；
• $B_i \bmod n$ 两两不同；
• 令 $s=\sum B_i-A_i$，则 $s \equiv 0 \pmod{m}$；
• $\max(B_i-A_i)\le \frac{s}{m}$。

这些条件是必要且充分的。

当 $s$ 固定，那么条件四越小越好，于是将 $A,B$ 取到排序不等式最小值更优。

有结论：最优解形如 $x,x+1,\cdots,x+n-1$。

证明考虑调整法，若 $B_n-B_1>n$，令 $B_1\gets B_n-n,B_n\gets B_1+n$。

则新的 $B$ 易证依旧满足四个条件，得证。

考虑最小化 $x$ 的值，由条件一和条件四可得 $x$ 的下界。

考虑条件三，暴力枚举 $n$ 个即可。

时间复杂度 $\mathcal O(n)$。

//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize("Ofast,fast-math")
//#pragma GCC target("avx,avx2")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 #define int long long
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef double db;
#define F(i, a, b) for(int i = a; i <= (b); ++i)
#define F2(i, a, b) for(int i = a; i < (b); ++i)
#define dF(i, a, b) for(int i = a; i >= (b); --i)
template<typename T> void debug(string s, T x) {
	cerr << "[" << s << "] = [" << x << "]\n";
}
template<typename T, typename... Args> void debug(string s, T x, Args... args) {
	for (int i = 0, b = 0; i < (int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++;
	else if (s[i] == ')' || s[i] == '}') b--;
	else if (s[i] == ',' && b == 0) {
		cerr << "[" << s.substr(0, i) << "] = [" << x << "] | ";
		debug(s.substr(s.find_first_not_of(' ', i + 1)), args...);
		break;
	}
}
#ifdef ONLINE_JUDGE
#define Debug(...)
#else
#define Debug(...) debug(#__VA_ARGS__, __VA_ARGS__)
#endif
#define pb push_back
#define fi first
#define se second
#define Mry fprintf(stderr, "%.3lf MB\n", (&Med - &Mbe) / 1048576.0)
#define Try cerr << 1e3 * clock() / CLOCKS_PER_SEC << " ms\n";
typedef long long ll;
// namespace Fread {const int SIZE = 1 << 17; char buf[SIZE], *S, *T; inline char getchar() {if (S == T) {T = (S = buf) + fread(buf, 1, SIZE, stdin); if (S == T) return '\n';} return *S++;}}
// namespace Fwrite {const int SIZE = 1 << 17; char buf[SIZE], *S = buf, *T = buf + SIZE; inline void flush() {fwrite(buf, 1, S - buf, stdout), S = buf;} inline void putchar(char c) {*S++ = c;if (S == T) flush();} struct NTR {~NTR() {flush();}} ztr;}
// #ifdef ONLINE_JUDGE
// #define getchar Fread::getchar
// #define putchar Fwrite::putchar
// #endif
inline int ri() {
	int x = 0;
	bool t = 0;
	char c = getchar();
	while (c < '0' || c > '9') t |= c == '-', c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
	return t ? -x : x;
}
inline void wi(int x) {
	if (x < 0) {
		putchar('-'), x = -x;
	}
	if (x > 9) wi(x / 10);
	putchar(x % 10 + 48);
}
inline void wi(int x, char s) {
	wi(x), putchar(s);
}
bool Mbe;
// mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3f;
const int _ = 3e5 + 5;

int n, m, a[_], sum, x;

bool Med;
signed main() {
	// Mry;
	n = ri(), m = ri();
	F(i, 1, n) a[i] = ri();
	sort(a + 1, a + n + 1);
	F(i, 1, n) x = max(x, a[i] - i + 1);
	F(i, 1, n) sum += (x + i - 1) - a[i];
	bool flg = 0;
	F(i, 0, n) if((sum + n * i) % m == 0) {
		x += i;
		sum += n * i;
		flg = 1;
		break;
	}
	if(!flg) return puts("-1"), 0;
	int mx = 0;
	F(i ,1, n) mx = max(mx, x + i - 1 - a[i]);
	while(mx > sum / m) {
		mx += m / __gcd(n, m), sum += n * m / __gcd(n, m);
	}
	cout << sum / m << '\n';
	// Try;
	return 0;
}