• <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
            久久综合噜噜激激的五月天| 国内精品久久久久久麻豆| 欧美精品九九99久久在观看| 久久精品国产亚洲5555| 久久亚洲日韩看片无码| 色诱久久久久综合网ywww| 久久精品国产精品亚洲精品| 国产成人综合久久久久久| 免费精品久久久久久中文字幕| 麻豆av久久av盛宴av| 丰满少妇人妻久久久久久| 精品久久久久久久久午夜福利| 久久99国产精品成人欧美| 久久香蕉国产线看观看精品yw| 久久久久久久综合日本| 狠狠色婷婷综合天天久久丁香| 色综合久久夜色精品国产| 久久精品成人免费观看97| 粉嫩小泬无遮挡久久久久久| 97视频久久久| 久久精品综合网| 伊人色综合九久久天天蜜桃| 国产精品永久久久久久久久久| 好久久免费视频高清| 婷婷五月深深久久精品| 97视频久久久| 久久精品国产免费观看三人同眠| 久久久艹| 四虎亚洲国产成人久久精品| 狠狠色综合网站久久久久久久| www.久久热| 91精品国产综合久久四虎久久无码一级| 人妻无码αv中文字幕久久| 99久久精品免费看国产一区二区三区| 久久人人爽人人爽人人片AV麻豆| 999久久久免费国产精品播放| 狠狠色噜噜狠狠狠狠狠色综合久久| 久久精品国产亚洲AV电影| …久久精品99久久香蕉国产| 国产精品一区二区久久国产| 91久久精一区二区三区大全|