浙江中医药大学暑期训练测试赛八A

本文介绍了一个基于 Naruto 使用影分身术的算法问题。目标是最小化使用技能后的疲劳值,并生成尽可能多的分身。通过递归算法求解最优解。

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1374: Greatest Naruto Army

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 117   Solved: 37
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Description

ccc is a VERY VERY BIG fans of Naruto(漩涡鸣人). Naruto's perseverance and passion attract him a lot, so does Naruto's skill "Kagebunsin no jyutu"(影分身术). Actually, ccc knows more details about "Kagebunsin no jyutu" than most of us. When Naruto makes an new illusion(幻象), his Chakra will divide into two parts A and B, and increase Naruto's weariness value by absolutely value of (A - B). 
For example, at first Naruto have 5 Chakra, then he makes an illusion who has 2 Chakra. So he left 3 Chakra, and increase his weariness value by 1 (3 - 2). If he make a new illusion again and give him 2 Chakra, then he only left 1 Chakra and his weariness value changes to 2. Naruto cannot make a new illusion unless his Chakra is bigger than 1. What's more, Naruto's illusion can also use "Kagebunsin no jyutu" and the weariness they gains will also return to Naruto himself.

Now Naruto wants to make as many illusions as he can because Kakashi Sensei wants to teach him a new Ninjyutsu(忍术) and minimum his weariness value.

Input

The first line of input contains an integer T, indicating the number of test cases, then T lines follow , each line contains a positive integer N, indicating the number of Naruto's Chakra.
0 < N < 10000000;

Output

For each case ,output the minimum weariness value of Naruto when he makes the most illusions.

Sample Input

259

Sample Output

23

AC代码:


#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
int min(int n)
{
        if(n==1||n==2||n==0)
                return0;
        if(n%2==0)
                returnmin(n/2)*2;
        else
                returnmin(n/2)+min(n/2+1)+1;
}
intmain()
{
        int t;
        scanf("%d",&t);
        while(t--)
        {
                long long n;
                scanf("%lld",&n);
                printf("%d\n",min(n));
        }
        return0;
}
 

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