D - D
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice Gym
100952D
Description
standard input/output
You have been out of Syria for a long time, and you recently decided to come back. You remember that you have M friends there and since you are a generous man/woman you want to buy a gift for each of them, so you went to a gift store that have N gifts, each
of them has a price.
You have a lot of money so you don‘t have a problem with the sum of gifts‘ prices that you‘ll buy, but you have K close friends among
your M friends you want their gifts to be expensive so the price of each of them is at least D.
Now you are wondering, in how many different ways can you choose the gifts?
Input
The input will start with a single integer T, the number of test cases. Each test case consists of two lines.
the first line will have four integers N, M, K, D (0 ?≤? N, M ?≤? 200, 0 ?≤? K ?≤? 50,
0 ?≤? D ?≤? 500).
The second line will have N positive integer number, the price of each gift.
The gift price is ?≤? 500.
Output
Print one line for each test case, the number of different ways to choose the gifts (there will be always one way at least to choose the gifts).
As the number of ways can be too large, print it modulo 1000000007.
Sample Input
Input
2 5 3 2 100 150 30 100 70 10 10 5 3 50 100 50 150 10 25 40 55 300 5 10
Output
3 126
Source
UESTC 2016 Summer Training #21
My Solution
组合学
有 n 个礼物， m个朋友， 其中k 个很要好的朋友， 要买 price 大于 d 的礼物
领 price >= d 的礼物个数为 cnt
则如果用 C[cnt][k] * C[n - k][m - k] 则显然不对， 因为这里面前面选的 price >= d的， 后面给普通好朋友选礼物的时候也会选到， 这样总的买的礼物数来说有大量重复了
所以 应该是分步 分类(price >= d 的与 < d 的分开算， 这样就不会相同的礼物选2次了)
首先 C[cnt][k] * C[n - cnt][m - k];
然后 C[cnt][k + 1] * C[n - cnt][ m - k - 1]
接着 C[cnt][k + 2] * C[n - cnt][ m - k - 2]
......
直到 k + 1 == cnt 或者 m - k - 1 < 0 //其中 如果k + 1 == cnt 则cnt 选完了，而 m - k - 1 < 0 则是 则是全都选了price >= d的了
复杂度 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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