原題傳送門
冷靜分析了一下,貌似很可做的樣子
數據範圍告訴我要用一個的做法
對於每個數,可以分類討論,是否把這個數翻倍
- 不翻倍,也不能翻倍,剩下的數假設有個,答案爲
- 翻倍,都必須翻倍,設都要翻倍的數有個,答案爲
如何求上面的?用二分就好了
Code:
#include <bits/stdc++.h>
#define maxn 100010
#define LL long long
using namespace std;
const LL qy = 998244353;
LL fac[maxn], inv[maxn], ans[maxn], a[maxn], b[maxn];
int n, k;
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
LL C(int n, int m){ return m > n ? 0 : fac[n] * inv[n - m] % qy * inv[m] % qy; }
LL pow(LL n, LL k){
if (!k) return 1;
LL sum = pow(n, k >> 1);
sum = sum * sum % qy;
if (k & 1) sum = sum * n % qy;
return sum;
}
int find1(int x){
int l = 1, r = n, ans = 0;
while (l <= r){
int mid = (l + r) >> 1;
if (b[mid] >= x) ans = mid, r = mid - 1; else l = mid + 1;
}
return ans ? ans : n;
}
int find2(int x){
int l = 1, r = n, ans = 0;
while (l <= r){
int mid = (l + r) >> 1;
if (b[mid] <= x) ans = mid, l = mid + 1; else r = mid - 1;
}
return ans ? ans : 0;
}
int main(){
n = read(), k = read();
fac[0] = 1;
for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % qy;
inv[n] = pow(fac[n], qy - 2);
for (int i = n - 1; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1) % qy;
for (int i = 1; i <= n; ++i) a[i] = b[i] = read();
sort(b + 1, b + 1 + n);
for (int i = 1; i <= n; ++i){
if (!a[i]){ ans[i] = C(n, k); continue; }
int l = find2(a[i] % 2 == 0 ? a[i] / 2 - 1 : a[i] / 2), r = find1(a[i]), x = n - (r - l);
ans[i] = C(x, k);
l = find1(a[i]), r = find2(2 * a[i] - 1), x = r - l + 1;
if (k >= x) (ans[i] += C(n - x, k - x)) %= qy;
}
for (int i = 1; i <= n; ++i) printf("%lld\n", ans[i]);
return 0;
}