UVA11384 Help is needed for Dexter【数学】

Dexter is tired of Dee Dee. So he decided to keep Dee Dee busy in a game. The game he planned for　her is quite easy to play but not easy to win at least not for Dee Dee. But Dexter does not have time　to spend on this silly task, so he wants your help.
    There will be a button, when it will be pushed a random number N will be chosen by computer. Then on screen there will be numbers from 1 to N. Dee Dee can choose any number of numbers from the numbers on the screen, and then she will command computer to subtract a positive number chosen by her (not necessarily on screen) from the selected numbers. Her objective will be to make all the numbers 0.
    For example if N = 3, then on screen there will be 3 numbers on screen: 1, 2, 3. Say she now selects 1 and 2. Commands to subtract 1, then the numbers on the screen will be: 0, 1, 3. Then she selects 1 and 3 and commands to subtract 1. Now the numbers are 0, 0, 2. Now she subtracts 2 from 2 and all the numbers become 0.
    Dexter is not so dumb to understand that this can be done very easily, so to make a twist he will give a limit L for each N and surely L will be as minimum as possible so that it is still possible to win within L moves. But Dexter does not have time to think how to determine L for each N, so he asks you to write a code which will take N as input and give L as output.
Input
Input consists of several lines each with N such that 1 ≤ N ≤ 1, 000, 000, 000. Input will be terminated by end of file.
Output
For each N output L in separate lines.
Sample Input
1
2
3
Sample Output
1
2
2

问题链接：UVA11384 Help is needed for Dexter
问题简述：（略）
问题分析：
    给定一个Ｎ，对１－Ｎ这Ｎ数，选出若干个数，对这些数都减去某个数，重复这样的操作，问最少几次后所有的数都变为０。
例如：
１　２　３　４　５
先对后３个，都减去３，则变为：
１　２　０　１　２
对于所有的２，都减去２，则变为：
１　０　０　０　１
再对于所有的１，都减去１，则变为：
０　０　０　０　０
    对于给定的Ｎ，先考虑起中值，中值右边所有数都减去中值，至少１个数已经变为０。然后从新排序，重复这个过程。这样可以得出所有数变为０需要做减法的次数是f(n)=1+f(n/2)。
    对以上的递推式，需要找出其规律，然后再打表搜索即可。对公式f(n)=1+f(n/2)，可以找出其中的规律，那就是换１次的有２^0（Ｎ=1）个，2次的有２^1（N=2-3）个，需要换３次有2^2（Ｎ=4-7）个，需要换４次有2^3（N=8-15）个，…
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA11384 Help is needed for Dexter */#include <bits/stdc++.h>using namespace std;//int f(int n)
//{//    if(n == 1) return 1;
//    else return 1 + f(n / 2);
//}const int N = 32;
int dp[N];void init()
{//    for(int i = 1; i <= 17; i++)
//        printf("%d %d\n", i, f(i));int t = 1;for(int i = 0; i < N; i++)dp[i] = t << i;
}int main()
{init();int n;while(~scanf("%d", &n))for(int i = 0; i < N; i++)if(n >= dp[i] && n <= dp[i + 1] - 1) {printf("%d\n", i + 1);}return 0;
}

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