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| 1 | +/* |
| 2 | +Problem: |
| 3 | +Pick k bloggers from n to maximize total followers, and count how many ways we can do this. |
| 4 | +The answer should be modulo 10^9 + 7. |
| 5 | +
|
| 6 | +Approach: |
| 7 | +To maximize the sum, we must pick bloggers with the highest follower counts. |
| 8 | +I sort the array in descending order. The "threshold" value will be the followers of the k-th blogger. |
| 9 | +I then count how many bloggers in the entire list have this threshold value and how many |
| 10 | +we actually need to pick from them to complete our k-blogger selection. |
| 11 | +The result is simply (total_with_threshold) Choose (needed_from_them). |
| 12 | +I used Pascal's triangle to precompute combinations since n is small (up to 1000). |
| 13 | +
|
| 14 | +TimeComplexity: O(n^2) precompute + O(n log n) per test case |
| 15 | +Spacecomplexity: O(nCr table size) approx O(n^2) |
| 16 | +
|
| 17 | +Submission Link: https://codeforces.com/contest/1475/submission/355409229 |
| 18 | +*/ |
| 19 | + |
| 20 | +#include <iostream> |
| 21 | +#include <vector> |
| 22 | +#include <algorithm> |
| 23 | + |
| 24 | +using namespace std; |
| 25 | + |
| 26 | +long long nCr[1005][1005]; |
| 27 | +const int MOD = 1e9 + 7; |
| 28 | + |
| 29 | +void pre() { |
| 30 | + for (int i = 0; i <= 1000; i++) { |
| 31 | + nCr[i][0] = 1; |
| 32 | + for (int j = 1; j <= i; j++) { |
| 33 | + nCr[i][j] = (nCr[i - 1][j - 1] + nCr[i - 1][j]) % MOD; |
| 34 | + } |
| 35 | + } |
| 36 | +} |
| 37 | + |
| 38 | +void solve() { |
| 39 | + int n, k; |
| 40 | + cin >> n >> k; |
| 41 | + |
| 42 | + vector<int> a(n); |
| 43 | + for (int i = 0; i < n; i++) cin >> a[i]; |
| 44 | + |
| 45 | + sort(a.begin(), a.end(), greater<int>()); |
| 46 | + |
| 47 | + int target = a[k - 1]; |
| 48 | + int cnt = 0; |
| 49 | + for (int x : a) { |
| 50 | + if (x == target) cnt++; |
| 51 | + } |
| 52 | + |
| 53 | + int take = 0; |
| 54 | + for (int i = 0; i < k; i++) { |
| 55 | + if (a[i] == target) take++; |
| 56 | + } |
| 57 | + |
| 58 | + cout << nCr[cnt][take] << endl; |
| 59 | +} |
| 60 | + |
| 61 | +int main() { |
| 62 | + ios::sync_with_stdio(0); |
| 63 | + cin.tie(0); |
| 64 | + |
| 65 | + pre(); |
| 66 | + |
| 67 | + int t; |
| 68 | + cin >> t; |
| 69 | + while (t--) { |
| 70 | + solve(); |
| 71 | + } |
| 72 | + |
| 73 | + return 0; |
| 74 | +} |
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