Gym 100952D time-to-go back combinatorics, Yang Hui triangle preprocessing combined number

D-DTime limit:MS Memory Limit:262144KB 64bit IO Format:%i64d &%i64u SubmitStatusPracticeGym 100952D

Description

Standard Input/output
Statements

You are been out of Syria for a long time, and you recently decided to come back. You remember this you had M friends there and since you is a generous man/woman you want to buy a gift for each of the them, So-went to a gift store, which has N-gifts, each of the them has a price.

You had a lot of money so you don ' t has a problem with the sum of gifts ' prices which you'll buy, but you have K close fr Iends among

Your M friends you want their gifts to being expensive so the price of each of the them are at least D.

Now is wondering, how do many different ways can you choose the gifts?

Input

The input would start with a single integer T, the number of test cases. Each test case consists of lines.

The first line'll has four integers N, M, K, D (0 ? ≤?). N, M ≤ 0, ≤? K . ≤ 0 ? ≤? D. ≤ 500).

The second line would have N positive integer number and the price of each gift.

The gift price is ? ≤ ? 500.

Output

Print one line for all test case, the number of different ways to choose the gifts (there'll be a always one-off at least To choose the gifts).

As the number of ways can too large, print it modulo 1000000007.

Sample Input

Input

25 3 2 100150 30 100 70 1010 5 3 50100 50 150 10 25 40 55 300 5 10

Output

3126

Source

UESTC Summer Training #21

Gym 100952D

My Solution

Combinatorial studies

There are n gifts, m friends, of whom K is a good friend, to buy a gift of price greater than D

The number of gifts for >= D is CNT

If use C[cnt][k] * C[n-k][m-k] is obviously wrong, because here before the choice of price >= D, the back to the ordinary good friend choose the gift of the time will also be selected, so the total number of gifts to buy a lot of repetition

So it should be sorted by step (price >= D and < D separately, so it won't be the same gift selected 2 times)

First C[cnt][k] * c[n-cnt][m-k];

Then C[CNT][K + 1] * c[n-cnt][m-k-1]

Next C[cnt][k + 2] * c[n-cnt][m-k-2]

 ......

Until K + 1 = = CNT or m-k-1 < 0//where if k + 1 = = CNT is selected, and M-k-1 < 0 are all selected price >= D

Complexity O (T * N)

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef Long Long ll;const int MAXN = 2*1e2 + 8;const ll Hash = 1000000007;inline ll mod (ll a) {return a-(A/hash) *hash;}    LL C[MAXN][MAXN], val[maxn];inline void GetC () {memset (c, 0, sizeof C);        for (int i = 0; i < MAXN; i++) {c[i][0] = 1;        for (int j = 1; J <= I; j + +) {C[i][j] = mod (c[i-1][j-1] + c[i-1][j]); }}}inline BOOL CMP (ll A, ll b) {return a > B;}    int main () {#ifdef LOCAL freopen ("A.txt", "R", stdin);    Freopen ("B.txt", "w", stdout);    #endif//LOCAL GetC ();    int T, N, M, K, D, CNT;    LL ans;    scanf ("%d", &t);        while (t--) {cnt = 0; ans = 0;        scanf ("%d%d%d%d", &n, &m, &k, &d);        for (int i = 0; i < n; i++) {scanf ("%i64d", &val[i]);        } sort (val, Val + N, CMP); for (int i = 0; i < n; i++) {if (Val[i] < D) Break        else cnt++;        }//cout<<cnt<<endl;        cout<<c[n-k][m-k]<<endl;            for (int i = k; I <= cnt; i++) {if (n-cnt = = 0) break;            if (M-i < 0) break;        ans = mod (ans + mod (c[cnt][i] * c[n-cnt][m-i));        } if (n-cnt = = 0) ans = c[cnt][m];        ans = mod (c[cnt][k] * c[n-k][m-k]);        if (CNT < K | | n < m) printf ("0\n");    else printf ("%i64d\n", ans); } return 0;}

Thank you!

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Gym 100952D time-to-go back combinatorics, Yang Hui triangle preprocessing combined number