The Problem
问题
Consider the following algorithm:
考虑下面的算法：
| 1 2 3 4 5 6 | input n print n if n = 1 then stop if n is odd then n <- 3n + 1 else n <- n / 2 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.
给定输入22，接下来会打印出的的数列是：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.)
猜想输入任何整输算法都会最终停止（到打印出1为止）。尽管算法很简单，但这个猜想是否正确仍不得而知。然而经过验证，算法对于0 < n < 1,000,000的所有整除n都成立（事实上还验证过更多的数）。
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.
对于给定的一个输入n，算法能够打印出数字（包括1）的数量是可以确定的。对于给定的n，这个数量称作n的“周期长度”。在上面的例子中，22的周期长度为16。
For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.
给定任意的两个数i和j，你要确定在[i, j]的范围内的所有数中，最长的周期长度是多少。
The Input
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.
输入一系列的整数对i和j，每一对整除独占一行。所有整除都小于1,000,000且大于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.
你要处理所有的整除对，并确定每对整除所限定范围内（含）的所有整数的最大周期长度。
You can assume that no operation overflows a 32-bit integer.
你可以假设不计算中不会出现32位整数的溢出错误。
The Output
输入
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).
对输入的每一对i和j应该输出i、j和[i, j]范围内的最大周期长度。这3个数字之间各由至少1个空格隔开，并在每一行输入之后，将这3个输全部输出在下一行内。输出整数i和j时必须按照他们在输入时的顺序输入，后面跟着输出最大周期长度（在同一行）。
Sample Input
输入示例
1 10
100 200
201 210
900 1000
Sample Output
输出示例
1 10 20
100 200 125
201 210 89
900 1000 174
分析：这个题目提交了几次，前面都是WA，后来网上分析i和j的大小可能不同，于是再加上一个判断即可，还有因为数据比较大的时候注意时间，尽量优化程序….
1 #include "iostream" 2 using namespace std; 3 unsigned int judge(int n); 4 int main(void) 5 { 6 int i,i1; 7 int j,j1; 8 //int len; 9 //int temp1; 10 unsigned int temp2; 11 int m=0; 12 unsigned int result; 13 while((scanf("%d %d",&i,&j))!=EOF) 14 { 15 if(i>j) 16 { 17 i1=j; 18 j1=i; 19 } 20 else 21 { 22 i1=i; 23 j1=j; 24 } 25 //len=i1-j1+1; 26 //temp1=i1; 27 temp2=judge(i1); 28 for(m=i1;m<=j1;m++) 29 { 30 result=judge(m); 31 if(result>temp2) 32 temp2=result; 33 } 34 35 cout<<i<<" "<<j<<" "<<temp2<<endl; 36 } 37 return 0; 38 } 39 unsigned int judge(int n) 40 { 41 unsigned int count=1; 42 while(n>1) 43 { 44 if(n%2==1) 45 n=3*n+1; 46 else n=n/2;//其实这里可以优化为 n=(n%2==1)?(3*n+1):(n/2) 47 count++; 48 } 49 50 return count; 51 }
转载于:https://www.cnblogs.com/kb342/p/3711479.html