Problem J
Candy
In the ancient city of Ica, there is said to be a palace with wealth beyond imagination. Inside, there is a corridor with $N$ boxes of candy from all over the world. Travellers passing by can take as much candy as they want, provided that they pay its weight in gold.
The boxes of candy are numbered $0$ to $N-1$ from left to right. In box $i$, there are $a_ i$ units of candy left, where $a_ i$ is a non-negative integer.
As the guardian of the palace, you would like to move the boxes around so that boxes with a lot of candy end up closer to the entrance.
You are given the array $a_0, a_1, \ldots , a_{N-1}$, as well as the numbers $F$ and $T$. In a single operation, you are allowed to swap two adjacent elements of $a_0, a_1, \ldots , a_{N-1}$. What is the minimum number of operations required so that the first $F$ elements of the array sum to at least $T$?
Input
The first line of the input contains three integers, $N$, $F$, and $T$.
The second line of the input contains $N$ integers $a_0, a_1, \ldots , a_{N-1}$.
Output
If it is impossible to achieve the objective using the operations, print NO.
Otherwise, print a single integer, the minimal number of operations.
Constraints and Scoring
-
$1 \le N \le 100$.
-
$1 \le F \le N$.
-
$0 \le T \le 10^{11}$.
-
$0 \le a_ i \le 10^9$ for $i = 0, 1, \ldots , N-1$.
Note: The numbers in the input may not fit in a $32$-bit integer, so be aware of overflows if you are using C++.
Your solution will be tested on a set of test groups, each worth a number of points. Each test group contains a set of test cases. To get the points for a test group you need to solve all test cases in the test group.
|
Group |
Score |
Limits |
|
1 |
6 |
$N \le 2$ and $a_ i \le 100$ for $i = 0, 1, \ldots , N-1$ and $T \le 10^9$ |
|
2 |
19 |
$a_ i \le 1$ for $i = 0, 1, \ldots , N-1$ |
|
3 |
16 |
$N \le 20$ |
|
4 |
30 |
$a_ i \le 100$ for $i = 0, 1, \ldots , N-1$ |
|
5 |
29 |
No additional constraints |
Example
In the first sample test case, the first two elements should sum to at least $27$. This can be achieved by a single swap of two adjacent elements: swap the $4$ and $20$. After this swap, the array becomes 10 20 4 6 3 3, and indeed the first two elements sum to $10+20 = 30\ge 27$.
In the second sample test case, the $0$ must move all the way to the end of the array; this takes three swaps.
In the third sample test case, it is impossible to make the first two elements sum to at least $100$; the best we can do is $60+30=90$.
| Sample Input 1 | Sample Output 1 |
|---|---|
6 2 27 10 4 20 6 3 3 |
1 |
| Sample Input 2 | Sample Output 2 |
|---|---|
6 5 5000000000 1000000000 1000000000 0 1000000000 1000000000 1000000000 |
3 |
| Sample Input 3 | Sample Output 3 |
|---|---|
3 2 100 20 30 60 |
NO |
| Sample Input 4 | Sample Output 4 |
|---|---|
1 1 100 100 |
0 |
