Loj 6432. 「PKUSC2018」真实排名 （组合数）

题面

Loj

题解

枚举每一个点

分两种情况

翻倍or不翻倍

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\(1.\)如果这个点\(i\)翻倍, 要保持排名不变，哪些必须翻倍，哪些可以翻倍？

必须翻倍： \(a[i] \leq a[x] < a[i]*2\)

那么其他的都可以选择性翻倍

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\(2.\) 考虑点\(i\)不翻倍，

不能翻倍的： \(a[i]/2 \leq a[x] < a[i]\)

注意有和\(a[i]\)相等的可以翻倍

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以上可以排序后，二分+组合数算

细节比较多，具体看代码

Code

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>

#define ll long long

using namespace std;

const int N = 100010;
const ll Mod = 998244353;
ll ksm(ll x, ll y) {
    ll s = 1;
    while (y) {
        if (y & 1) s = s * x % Mod;
        y >>= 1;
        x = x*x%Mod;
    }
    return s;
}

ll fac[N];

ll C(int n, int m) {
    if (n < m || n < 0 || m < 0) return 0;
    return fac[n]*ksm(fac[m], Mod-2)%Mod*ksm(fac[n-m], Mod-2)%Mod;
}

int a[N], b[N];

template<class T> inline void read(T &x) {
    x = 0;  char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}

int main() {

    int n, k;
    read(n); read(k);
    for (int i = 1; i <= n; i++) read(a[i]), b[i] = a[i];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = fac[i-1]*i%Mod;
    
    sort(b+1, b+1+n);
    
    for (int i = 1; i <= n; i++) {

        if (!a[i]) {
            printf("%lld\n", C(n, k));
            continue;
        }
        
        ll ans1 = 0, ans2 = 0;
        
        int x1 = lower_bound(b+1, b+1+n, (a[i]+1)/2) - b - 1;
        int x2 = lower_bound(b+1, b+1+n, a[i]) - b;
        ans1 = C(x1+n-x2, k);

        int x3 = lower_bound(b+1, b+1+n, a[i]*2) - b;
        
        ans2 = C(n - x3 + x2, k - x3 + x2);

        printf("%lld\n", (ans1 + ans2) % Mod);      
    }   
    return 0;
}