首页 > ACM题库 > HDU-杭电 > HDU 1032 The 3n + 1 problem-模拟-[解题报告] C++
2013
11-28

HDU 1032 The 3n + 1 problem-模拟-[解题报告] C++

The 3n + 1 problem

问题描述 :

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.

Consider the following algorithm:

    1.      input n

    2.      print n

    3.      if n = 1 then STOP

    4.           if n is odd then n <- 3n + 1

    5.           else n <- n / 2

    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.

输入:

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.

输出:

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).

样例输入:

1 10
100 200
201 210
900 1000

样例输出:

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

 

直接模拟,超时了几次。

GCC的输入应该是while(scanf(“%d %d”, &a, &b) != EOF), 我写成了while(scanf(“%d %d”, &a, &b))一直超时。= =!

#include <stdio.h>

int s(int i) {
    int n = 1;
    while(i > 1) {
        if (i%2 == 0)
            i /= 2;
        else
            i = 3*i+1;
        n++;
    }
    return n;
}


int main() {
    int a, b;
    int n, m;
        
    while(scanf("%d %d", &n, &m) != EOF) {
        a = n;
        b = m;
        if (a > b) {
            int t=a;
            a=b;
            b=t;
        }
        int sum = 0;
        int i, r;
        for (i = a; i <= b; i++) {
            r = s(i);
            if (r > sum)
                sum = r;
        }
        
        printf("%d %d %d\n", n, m, sum);

                                
    }
    return 0;
}

  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  2. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮