• <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(1309) 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
            岛国搬运www久久| 亚洲成色999久久网站| 久久婷婷国产剧情内射白浆| 久久精品水蜜桃av综合天堂| 久久99精品久久久久久野外| 2021久久国自产拍精品| 久久久久免费视频| 久久精品国产亚洲沈樵| 麻豆亚洲AV永久无码精品久久| 亚洲国产一成久久精品国产成人综合 | 99re久久精品国产首页2020| 99久久国产主播综合精品| 久久久久久国产精品免费免费| 欧洲人妻丰满av无码久久不卡 | 久久人人爽人人爽人人片av麻烦 | 国产AⅤ精品一区二区三区久久| 要久久爱在线免费观看| 国产成人精品久久免费动漫| 久久久久黑人强伦姧人妻| 四虎国产精品成人免费久久| 久久影院亚洲一区| 国产激情久久久久影院小草| 亚洲AV日韩精品久久久久| 亚洲va国产va天堂va久久| 久久精品夜色噜噜亚洲A∨| 曰曰摸天天摸人人看久久久| 久久久久亚洲AV无码网站| 久久精品人妻中文系列| 四虎国产精品成人免费久久| 久久精品这里只有精99品| 伊人久久综合热线大杳蕉下载| 久久99精品久久只有精品| 久久久久久免费一区二区三区| 天天久久狠狠色综合| av国内精品久久久久影院| 潮喷大喷水系列无码久久精品| 香蕉久久夜色精品升级完成 | 久久精品国产国产精品四凭| 久久综合丝袜日本网| 久久久久国色AV免费看图片| 久久久久久午夜精品|