用c++写Bob has a magic computer which has infinite USB ports. But there is a problem with this computer——it can only read and write the two most recently inserted USB disks. Initially, Bob has n USB disks and every USB disk has one unique file. He wants to insert all the n USB disks into his computer. Since his computer is old, he must insert USB disk in the following way: Select 2 unplugged USB disks, insert them into computer. The two USB disks inserted in the previous step will share the files with each other, which means after that both two USB disks will have all files either in the first one or in the second one. If all the n USB disks are in the computer, the task is finished. Otherwise, choose one USB disk in the computer, unplug it and go to step 1 . Bob wants to know the minimum number of n such that there is a sequence of operations that ends up with at least k different files in each USB disk. Input Only one number,k (2≤k≤100000 ),meaning that finally each USB disk owns at least k different files. Output You need to output one number n , which means the minimum number of USB disks in module 998244353.
时间: 2024-01-29 07:04:35 浏览: 75
以下是 C++ 代码实现:
```cpp
#include <iostream>
using namespace std;
const int MOD = 998244353;
int power(int a, int b) {
int res = 1;
while (b) {
if (b & 1) {
res = 1ll * res * a % MOD;
}
a = 1ll * a * a % MOD;
b >>= 1;
}
return res;
}
int main() {
int k;
cin >> k;
int ans = power(2, k - 1);
cout << ans << endl;
return 0;
}
```
我们可以发现,每次插入两个 USB 磁盘,它们的文件集合会并集,也就是说最后所有的磁盘的文件集合是相同的。因此,我们只需要考虑一张磁盘最终有多少个文件即可。又因为每次插入两个磁盘的操作相当于对两个磁盘的文件集合求并集,所以每次插入两个磁盘后,每个磁盘的文件数至少翻倍。因此,我们只需要让 $2^{n-1}$ 大于等于 $k$ 即可,得到答案 $n=\lceil\log_2k\rceil+1$。注意要对 $998244353$ 取模。
时间复杂度为 $O(1)$。
相关问题
Write a few lines of C++ code to show an example of an infinite loop. Explain why it is an infinite loop
Here's an example of an infinite loop in C++:
```c++
while(true){
// code that never exits the loop
}
```
In this example, the `while` loop condition is always `true`, which means the loop will never exit. The loop will continue to execute the block of code inside the curly braces indefinitely.
This is an infinite loop because the loop condition is always true and there is no way for the loop to terminate on its own. The only way to exit the loop is to use a `break` statement or an external event (such as terminating the program).
Write a generator function merge that takes in two infinite generators a and b that are in increasing order without duplicates and returns a generator that has all the elements of both generators, in increasing order, without duplicates.
Sure! Here's an implementation of the `merge` generator function:
```python
def merge(a, b):
ia, ib = iter(a), iter(b)
va, vb = next(ia), next(ib)
while True:
if va < vb:
yield va
va = next(ia)
elif vb < va:
yield vb
vb = next(ib)
else:
yield va
va, vb = next(ia), next(ib)
```
This generator function takes two infinite generators `a` and `b` that are in increasing order without duplicates, and returns a generator that has all the elements of both generators, also in increasing order without duplicates.
The function first initializes two iterators `ia` and `ib` for the generators `a` and `b`, respectively. It then initializes the variables `va` and `vb` with the first elements of `a` and `b`, respectively.
The function then enters an infinite loop that repeatedly compares the values of `va` and `vb` and yields the smaller value. If `va` is smaller than `vb`, the function yields `va` and advances the `ia` iterator to the next element of `a`. If `vb` is smaller than `va`, the function yields `vb` and advances the `ib` iterator to the next element of `b`. If `va` and `vb` are equal, the function yields `va` (since we want to remove duplicates) and advances both the `ia` and `ib` iterators to the next elements of `a` and `b`, respectively.
Here's an example usage of this generator function:
```python
>>> def gen1():
... i = 0
... while True:
... yield i
... i += 2
...
>>> def gen2():
... i = 1
... while True:
... yield i
... i += 2
...
>>> merged = merge(gen1(), gen2())
>>> [next(merged) for _ in range(10)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
```
This example defines two infinite generators `gen1` and `gen2` that generate all even and odd numbers, respectively. The `merge` function is then used to merge these two generators into a single generator `merged` that generates all the numbers in increasing order, without duplicates. Finally, the list comprehension `[next(merged) for _ in range(10)]` is used to consume the first 10 elements of `merged` and return them as a list. The result is `[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]`, which shows that `merged` generates all the even and odd numbers up to 9 in increasing order without duplicates.