• <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
            久久人人爽人人爽人人片av高请| 国产亚洲综合久久系列| 色综合久久久久综合99| 久久天天躁狠狠躁夜夜2020一| 无码人妻久久一区二区三区| 久久不射电影网| 日韩AV毛片精品久久久| 99久久99久久久精品齐齐| 国产—久久香蕉国产线看观看| 久久久www免费人成精品| 大蕉久久伊人中文字幕| 久久99热这里只有精品国产| 国产巨作麻豆欧美亚洲综合久久| 国产毛片欧美毛片久久久| 99久久免费只有精品国产| 人妻精品久久无码专区精东影业| 99久久免费只有精品国产| 久久久久久午夜成人影院| 青青草原综合久久大伊人导航| 久久久九九有精品国产| 精品久久久久久中文字幕大豆网| 久久久WWW免费人成精品| 99国产精品久久| 久久精品99久久香蕉国产色戒 | 国产精品美女久久久久AV福利| 成人久久免费网站| 亚洲欧美国产日韩综合久久| 激情久久久久久久久久| 91久久精品国产免费直播| 成人久久综合网| 国产午夜福利精品久久2021| 无码伊人66久久大杳蕉网站谷歌| 一级A毛片免费观看久久精品| 久久久精品日本一区二区三区| 久久亚洲国产午夜精品理论片| 国内精品伊人久久久久| 久久亚洲私人国产精品| 精品人妻久久久久久888| 国产69精品久久久久777| 亚洲一区二区三区日本久久九| 精品一区二区久久久久久久网站|