• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            技術,瞎侃,健康,休閑……

            mahu@cppblog 人類的全部才能無非是時間和耐心的混合物
            posts - 11, comments - 13, trackbacks - 0, articles - 12
              C++博客 :: 首頁 :: 新隨筆 :: 聯系 :: 聚合  :: 管理

            The 3n + 1 problem

            Posted on 2006-06-10 00:41 mahudu@cppblog 閱讀(1311) 評論(3)  編輯 收藏 引用 所屬分類: C/C++

            Background

            Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

            The Problem

            Consider the following algorithm:

            1.	input n

            2. print n

            3. if n = 1 then STOP

            4. if n is odd then tex2html_wrap_inline44

            5. else tex2html_wrap_inline46

            6. GOTO 2

            Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

            It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)

            Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.

            For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

            The Input

            The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.

            You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.

            You can assume that no opperation overflows a 32-bit integer.

            The Output

            For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

            Sample Input

            1 10
            100 200
            201 210
            900 1000

            Sample Output

            1 10 20
            100 200 125
            201 210 89
            900 1000 174

            Solution ?

            #include <iostream>

            using namespace std;

            ?

            int cycle(intm)

            {

            ?? int i = 1;

            ?? while (m != 1){

            ????? if(m%2)

            ??????? m = m*3 + 1;

            ????? else

            ??????? m /= 2;

            ????? i++;

            ?? }

            ?? return i;

            }??

            ?

            int main()

            {

            ?? int m,n,max,temp;

            ?? int mOriginal,nOriginal;

            ?? int i;

            ?

            ?? while (cin >> m >> n){

            ????? mOriginal = m;

            ????? nOriginal = n;

            ????? if (m > n){

            ??????? temp = m;

            ??????? m = n;

            ??????? n = temp;

            ????? }

            ?

            ????? max = cycle(m);

            ????? for (i = m+1; i <= n; i++){

            ??????? temp = cycle(i);

            ??????? if (temp > max){

            ?????????? max = temp;

            ??????? }

            ????? }?

            ????? cout << mOriginal << " " << nOriginal << " " << max << endl;

            ?? }

            ?? return 0;

            }

            Feedback

            # re: The 3n + 1 problem  回復  更多評論   

            2007-11-14 20:34 by 無意中看到
            你這個程序屬于很難通過的,基本上會碰到超時問題
            輸入: 1 1000000
            看你多牛的計算機3秒能搞出來
            這是典型的dp問題,暴力是不好用的

            # re: The 3n + 1 problem  回復  更多評論   

            2008-10-18 17:17 by TaiwanNo.1
            /*
            這個可以以0.7 sec完成
            */
            #include <stdio.h>

            int compute(int a)
            {
            int cnt = 1;
            while(a > 1)
            {
            a & 0x01 ? (a = (a<<1) + a + 1) : (a >>= 1);
            ++cnt;
            }
            return cnt;
            }


            int a, b, c;
            int i, j;
            int main(void)
            {

            while(0 < scanf("%d %d", &i, &j))
            {
            c = 0;
            i < j ? (a = i, b = j) : (a = j, b = i);
            while(a <= b)
            {
            int tmp = compute(a++);
            if(tmp > c)
            c = tmp;
            }
            printf("%d %d %d\n", i, j, c);
            }
            return 0;
            }

            # re: The 3n + 1 problem  回復  更多評論   

            2011-01-16 13:39 by UDHeart
            @TaiwanNo.1
            ...我就是這樣寫的,沒有這么快
            久久国产精品免费| 亚洲综合久久久| 婷婷久久久亚洲欧洲日产国码AV| 久久午夜无码鲁丝片午夜精品| 91精品国产色综久久| 国产成人综合久久精品尤物| 岛国搬运www久久| 青青草原综合久久大伊人导航| 人妻无码精品久久亚瑟影视| 欧美日韩精品久久久久| 性高湖久久久久久久久AAAAA| 亚洲欧美国产日韩综合久久| 思思久久精品在热线热| 亚洲午夜久久久久妓女影院 | 久久精品国产亚洲Aⅴ蜜臀色欲| 亚洲国产成人久久综合一| 狠狠人妻久久久久久综合蜜桃| 一本久久综合亚洲鲁鲁五月天亚洲欧美一区二区 | 久久国产精品99精品国产987| 久久精品国产影库免费看| 国产日韩欧美久久| 影音先锋女人AV鲁色资源网久久 | 香蕉久久一区二区不卡无毒影院| 久久精品免费网站网| 99精品国产免费久久久久久下载 | 久久久久亚洲爆乳少妇无| 免费久久人人爽人人爽av| .精品久久久麻豆国产精品| 久久人搡人人玩人妻精品首页| 欧美一区二区三区久久综| 国产成人精品久久亚洲高清不卡 | 久久久国产99久久国产一| 精品久久人妻av中文字幕| 久久夜色精品国产| 久久99精品久久久久久| yy6080久久| 久久成人18免费网站| 97热久久免费频精品99| 精品久久久久久中文字幕大豆网| 婷婷久久综合九色综合98| 亚洲乱码精品久久久久..|