A. Packets（数论小知识，1, 2, 4, .... , 2^n可以组成2^(n+1)

本文探讨了一种算法问题，即如何用最少数量的包来组合成任意从1到给定数值的所有整数。通过巧妙利用数论知识，文章提供了一个高效的解决方案，并附带代码实现，展示了如何快速找到满足条件的最小包数。

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A. Packets
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You have n
coins, each of the same value of 1

Distribute them into packets such that any amount x
(1≤x≤n

) can be formed using some (possibly one or all) number of these packets.

Each packet may only be used entirely or not used at all. No packet may be used more than once in the formation of the single x
, however it may be reused for the formation of other x

’s.

Find the minimum number of packets in such a distribution.
Input

The only line contains a single integer n
(1≤n≤109

) — the number of coins you have.
Output

Output a single integer — the minimum possible number of packets, satisfying the condition above.
Examples
Input
Copy

Output
Copy

Input
Copy

Output
Copy

Note

In the first example, three packets with 1
, 2 and 3 coins can be made to get any amount x (1≤x≤6

To get 1

use the packet with 1
coin.
To get 2
use the packet with 2
coins.
To get 3
use the packet with 3
coins.
To get 4
use packets with 1 and 3
coins.
To get 5
use packets with 2 and 3
coins
To get 6

use all packets.

In the second example, two packets with 1
and 1 coins can be made to get any amount x (1≤x≤2).

题目大意

用最少N可以组成1～X所有数，输入一个X求N。
这里使用了一个数论小知识，1,2,4，….，2^n可以组成2^(n+1) - 1所有数。
所以这道题，可以先用1,2,4，这些2^n数来选，然后剩下用一个数就可以填充了。

代码

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    int now = 0;
    int ans = 0;
    scanf("%d", &n);
    for(int i = 1; i < n; i*=2) {
        now += i;
        ans++;
    }
    if(now < n) ans++;
    cout << ans << endl;
    return 0;
}