【2020年杭电暑假第五场】6825 Set1

【2020年杭电暑假第五场】6825 Set1 组合数学+数学推导 / dp

    • 题意
    • 思路
      • 方法一:组合数学+数学推导
      • 方法二:dp
    • Code(3026MS)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=6825

题意

You are given a set S={1…n}. It guarantees that n is odd. You have to do the following operations until there is only 1 element in the set:

Firstly, delete the smallest element of S. Then randomly delete another element from S.

For each i∈[1,n], determine the probability of i being left in the S.

It can be shown that the answers can be represented by PQ, where P and Q are coprime integers, and print the value of P×Q−1 mod 998244353.

Input
The first line containing the only integer T(T∈[1,40]) denoting the number of test cases.

For each test case:

The first line contains a integer n .

It guarantees that: ∑n∈[1,5×106].

Output
For each test case, you should output n integers, i-th of them means the probability of i being left in the S.

Sample Input
1
3

Sample Output
0 499122177 499122177

给出一个 n n n,有 [ 1 , n ] [1,n] [1,n] n n n个数。
每次有如下操作:先删去最小的元素,再随机删掉一个元素。
求每个数留下来的概率期望。

思路

方法一:组合数学+数学推导

考虑元素 i i i留下来的方案数,那么他前面有 i − 1 i-1 i1个数,后面有 n − i n-i ni个数。
我们易得当 n − i ≥ i − 1 n-i\geq i-1 nii1时,i才会被留下来,即 i ≤ n 2 i\leq\frac{n}{2} i2n时都是 0 0 0,往后才有值。所以我们可以直接输出前 n 2 \frac{n}{2} 2n 0 0 0

然后在处理后面的数字。
我们设每次删除最小元素为操作一,随机删除元素为操作二。
i i i后面的 n − i n-i ni个数一定是操作二删除的,所以第一步,我们让后面 n − i n-i ni个数与前面 i − 1 i-1 i1中的 n − i n-i ni个一一对应删除就行。(在 i − 1 i-1 i1中选 n − i n-i ni个数出来在给他排序)
设第 i i i个数被留下来的方案数为 c n t [ i ] cnt[i] cnt[i],则
c n t [ i ] = C i − 1 n − i ∗ A n − i n − i = ( i − 1 ) ! ( 2 i − n − 1 ) ! cnt[i] = C_{i-1}^{n-i}*A_{n-i}^{n-i} = \frac{(i-1)!}{(2i-n-1)!} cnt[i]=Ci1niAnini=(2in1)!(i1)!

然后在考虑剩下的 ( i − 1 ) − ( n − i ) = ( 2 i − n − 1 ) (i-1)-(n-i)=(2i-n-1) (i1)(ni)=(2in1)个数,根据前面,在其中选择两两删除。
( 2 i − n − 1 ) (2i-n-1) (2in1)两两选择有 C 2 i − n − 1 2 ∗ C 2 i − n − 3 2 . . . C 4 2 ∗ C 2 2 C_{2i-n-1}^2*C_{2i-n-3}^2...C_4^2*C_2^2 C2in12C2in32...C42C22
( 2 i − n − 1 ) ! 2 ! ∗ ( 2 i − n − 3 ) ! ∗ ( 2 i − n − 3 ) ! 2 ! ∗ ( 2 i − n − 5 ) ! ∗ . . . ∗ 2 ! 2 ! ∗ 0 ! \frac{(2i-n-1)!}{2!*(2i-n-3)!}*\frac{(2i-n-3)!}{2!*(2i-n-5)!}*...*\frac{2!}{2!*0!} 2!(2in3)!(2in1)!2!(2in5)!(2in3)!...2!0!2!
( 2 i − n − 1 ) ! ( 2 ! ) 2 i − n − 1 2 \frac{(2i-n-1)!}{(2!)^{\frac{2i-n-1}{2}}} (2!)22in1(2in1)!

到这里还没有结束,然后我想了好久,是我太捞了。
比如 [ 1 , 6 ] [1,6] [1,6],我们选择的时候会重复选择多次,比如
( 1 , 2 ) ( 3 , 4 ) ( 5 , 6 ) (1,2)(3,4)(5,6) (1,2)(3,4)(5,6)
( 1 , 2 ) ( 5 , 6 ) ( 3 , 4 ) (1,2)(5,6)(3,4) (1,2)(5,6)(3,4)
( 3 , 4 ) ( 1 , 2 ) ( 5 , 6 ) (3,4)(1,2)(5,6) (3,4)(1,2)(5,6)
( 3 , 4 ) ( 5 , 6 ) ( 1 , 2 ) (3,4)(5,6)(1,2) (3,4)(5,6)(1,2)
( 5 , 6 ) ( 1 , 2 ) ( 3 , 4 ) (5,6)(1,2)(3,4) (5,6)(1,2)(3,4)
( 5 , 6 ) ( 3 , 4 ) ( 1 , 2 ) (5,6)(3,4)(1,2) (5,6)(3,4)(1,2)
这些都是上面包括的情况,因为选的先后次序不同,所以会出现重复的情况,一共出现相同的组合 ( n 2 ) ! (\frac{n}{2})! (2n)!,所以我们要在上面的基础上再除以 ( 2 i − n − 1 2 ) ! (\frac{2i-n-1}{2})! (22in1)!
那么剩下数 ( 2 i − n − 1 ) (2i-n-1) (2in1)中两两选择的方案数为
( 2 i − n − 1 ) ! ( 2 ! ) 2 i − n − 1 2 ∗ ( 2 i − n − 1 2 ) ! \frac{(2i-n-1)!}{(2!)^{\frac{2i-n-1}{2}}*(\frac{2i-n-1}{2})!} (2!)22in1(22in1)!(2in1)!
最后整合上述所有情况,第 i i i个数被留下的总方案数为
c n t [ i ] = ( i − 1 ) ! ( 2 i − n − 1 ) ! ∗ ( 2 i − n − 1 ) ! ( 2 ! ) 2 i − n − 1 2 ∗ ( 2 i − n − 1 2 ) ! cnt[i] = \frac{(i-1)!}{(2i-n-1)!}*\frac{(2i-n-1)!}{(2!)^{\frac{2i-n-1}{2}}*(\frac{2i-n-1}{2})!} cnt[i]=(2in1)!(i1)!(2!)22in1(22in1)!(2in1)!
化简得
c n t [ i ] = ( i − 1 ) ! ( 2 ! ) 2 i − n − 1 2 ∗ ( 2 i − n − 1 2 ) ! cnt[i] = \frac{(i-1)!}{(2!)^{\frac{2i-n-1}{2}}*(\frac{2i-n-1}{2})!} cnt[i]=(2!)22in1(22in1)!(i1)!

总方案数为 s u m = ∑ i = 1 n c n t [ i ] sum=\sum_{i=1}^ncnt[i] sum=i=1ncnt[i]
i i i个数被留下的概率期望为 p [ [ i ] = c n t [ i ] s u m p[[i]=\frac{cnt[i]}{sum} p[[i]=sumcnt[i]
时间复杂度为 O ( n ) O(n) O(n)

事先预处理所有的阶乘和阶乘逆元。

方法二:dp

参考:https://www.cnblogs.com/luyouqi233/p/13434815.html
dp方法我也不太会,哭了。

Code(3026MS)

#include 

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;

#define INF 0x7f7f7f
#define mem(a, b) memset(a , b , sizeof(a))
#define FOR(i, x, n) for(int i = x;i <= n; i++)

const ll mod = 998244353;
// const int maxn = 1e5 + 10;
// const double eps = 1e-6;

const int N = 5e6 + 5;

ll F[N];
ll invn[N];
ll invF[N];
ll cnt[N];

ll quick_pow(ll a, ll b)
{
    ll ans = 1;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans % mod;
}

void Init()
{
    F[0] = F[1] = invn[0] = invn[1] = invF[0] = invF[1] = 1;
    for(int i = 2;i < N; i++){
        F[i] = F[i - 1] * i % mod;
        invn[i] = (mod - mod / i) * invn[mod % i] % mod;
        invF[i] = invF[i - 1] * invn[i] % mod;
    }
}

void solve() {

    Init();

    int T;
    scanf("%d",&T);

    ll inv2 = 499122177;

    while(T--)
    {
        int n;
        scanf("%d",&n);
        if(n == 1) printf("1\n");
        else{
            printf("0");
            for(int i = 2;i <= n / 2; i++)
                printf(" 0");

            ll ans = 0;

            for(int i = n / 2 + 1;i <= n; i++) {
                cnt[i] = F[i - 1] * invF[(2 * i - n - 1) / 2] % mod * quick_pow(inv2, (2 * i - n - 1) / 2) % mod;
                ans = (ans + cnt[i]) % mod;
            }

            ll sum = quick_pow(ans, mod - 2);

            for(int i = n / 2 + 1;i <= n; i++) {
                printf("% lld",sum * cnt[i] % mod);
            }

            printf("\n");

        }
    }
}


signed main() {
    ios_base::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr);
#ifdef FZT_ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}

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