Javabeans（递推）

Javabeans

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Javabeans are delicious. Javaman likes to eat javabeans very much.

Javaman has n boxes of javabeans. There are exactly i javabeans in the i-th box (i = 1, 2, 3,...n). Everyday Javaman chooses an integer x. He also chooses several boxes where the numbers of javabeans are all at least x. Then he eats x javabeans in each box he has just chosen. Javaman wants to eat all the javabeans up as soon as possible. So how many days it costs for him to eat all the javabeans?

Input

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

Each test case is a line of a positive integer 0 < n < 2³¹.

Output

For each test case output the result in a single line.

Sample Input

4
1
2
3
4

Sample Output

1
2
2
3

这题的关键是x怎么选取，最优解取x = [n/2].

用f(k)表示n=k时所需的天数。

当 n=2k+1, k为非负整数，原序列为<1, 2, ..., k+1, ..., 2k+1>.中间的一个数是k+1，则先取 x=k+1，原序列变为<1, 2, ..., k , 0, 1, 2..., k>.可以发现，两边对称，于是 f(2k+1)=1+f(k).即f(n)=1+f([n/2]).

当n=2k, k为非负整数，原序列为<1, 2, ..., k, k + 1, ..., 2k>.取x=n/2=k，原序列变为<1, 2,  ..., k-1, 0 , 1, 2,..,  k>.两边对称，于是f(2k)=1+f(k).也满足f(n)=1+f([n/2]).

于是得到递推公式：f(n)=1+f([n/2]).

那么为什么要取x = [n/2]? 因为这能使得原序列“最快”地被分割为尽可能小的序列。不断取x = [n/2]就将原序列不断地二分。如果取x=[n/a].a>2,则分割成的序列中总有长度大于[n/2]的。

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#define LL long long
#define MAXI 2147483647
#define MAXL 9223372036854775807
#define eps (1e-8)
#define dg(i) cout << "*" << i << endl;
  
using namespace std;
  
int Solve(LL n)
{
    if(n == 1) return 1;
    return 1 + Solve(n / 2);
}
  
int main()
{
    LL n;  //尽管n是32位整型，但为免运算过程中发生溢出，仍设为64位
    int t;
    cin >> t;
    while(t--)
    {
        cin >> n;
        cout << Solve(n) << endl;
    }
    return 0;
}