「PKUSC2018」真实排名

题面

题解

因为操作为将一些数字翻倍，

所以对于一个数\(x\)，

能影响它的排名的的只有满足\(2y\geq x\)或\(2x>y\)的\(y\)

将选手的成绩排序，然后考虑当前点的方案

1. 不翻倍

此时，如果要保证\(x\)的排名不变，那么所有满足\(2y \geq x\)的\(y\)都不能动

设满足\(2y \geq x\)的数有\(\mathrm{Len}\)个，则方案数为\(\binom{n-\mathrm{Len}}{k}\)

2. 翻倍

此时，如果要保证\(x\)的排名不变，那么所有满足\(2x > y\)的\(y\)都要动

设满足\(2x > y\)的数有\(\mathrm{Len}\)个，则方案数为\(\binom{n-\mathrm{Len}}{k-\mathrm{Len}}\)

如果有相同的怎么办？？？

去重之后，将\(0\)和其他的数字分开处理即可。

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<vector>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(1e5 + 10), Mod(998244353), N(maxn - 10);
struct node { int val, id, cnt; } A[maxn], B[maxn];
inline bool cmp(const node &lhs, const node &rhs) { return lhs.val < rhs.val; }

int fastpow(int x, int y)
{
	int ans = 1;
	while(y)
	{
		if(y & 1) ans = 1ll * ans * x % Mod;
		x = 1ll * x * x % Mod, y >>= 1;
	}
	return ans;
}

int sum[maxn];
int fac[maxn], inv[maxn];
int n, K, ans[maxn], tot;
std::vector<int> S[maxn];

inline int C(int n, int m)
{
	if(n < m || n < 0 || m < 0) return 0;
	return 1ll * fac[n] * inv[m] % Mod * inv[n - m] % Mod;
}

int main()
{
	fac[0] = inv[0] = 1;
	for(RG int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
	inv[N] = fastpow(fac[N], Mod - 2);
	for(RG int i = N - 1; i; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % Mod;
	n = read(), K = read();
	for(RG int i = 1; i <= n; i++) B[i] = (node) {read(), i, 1};
	std::sort(B + 1, B + n + 1, cmp); B[0] = (node) {-1, -1, 0};
	for(RG int i = 1; i <= n; i++)
		if(B[i].val != B[i - 1].val)
			{ ++tot; A[tot] = B[i]; S[tot].push_back(B[i].id); }
		else ++A[tot].cnt, S[tot].push_back(B[i].id);
	for(RG int i = 1; i <= tot; i++)
		sum[i] = sum[i - 1] + A[i].cnt;
	for(RG int i = 1, R = 1; i <= tot; i++)
	{
		while(R < tot && A[i].val * 2 > A[R + 1].val) ++R;
		int Len = sum[R] - sum[i - 1];
		if(A[i].val == 0) Len = sum[R] - sum[i] + 1;
		if(Len <= K)
			for(RG int j = 0, l = S[i].size(); j < l; j++)
				ans[S[i][j]] = (ans[S[i][j]] + C(n - Len, K - Len)) % Mod;
	}
	for(RG int i = tot, L = tot; i; i--)
	{
		while(L > 1 && A[L - 1].val * 2 >= A[i].val) --L;
		int Len = sum[i - 1] - sum[L - 1] + 1;
		for(RG int j = 0, l = S[i].size(); j < l; j++)
			ans[S[i][j]] = (ans[S[i][j]] + C(n - Len, K)) % Mod;
	}
	for(RG int i = 1; i <= n; i++) printf("%d\n", ans[i]);
	return 0;
}