HDU 1032 The 3n+1 Problem
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The 3n + 1 problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem Description
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 n2. print n3. if n = 1 then STOP4. if n is odd then n <- 3n + 15. else n <- n / 26. 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.
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.
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
Source
UVA
问题如下:(1)输入一个正整数n;(2)如果n=1则结束;(3)如果n是奇数,则n变为3n+1,否则n变为n/2;(4)转入第(2)步。(5)输出最长序列的数(从n算到1)
10:5 16 8 4 2 19:28 14 7 22 11 34 17 52 26~8:4 2 17:22 11 34 17 52 26 13 40 20~ 6:3 10 5 16 8 4 2 15:16 8 4 2 14:2 13:10 5 16 8 4 2 12:11:0 **************************10→79→208→47→176→95→64→3 3→8 2→2 1→1
#include<stdio.h>#include<math.h>int main(){ int i,j,k,n,max=0,temp; while(scanf("%d %d",&i,&j)!=EOF)、 { printf("%d %d ",i,j); //按输入格式顺序输出 if(i>j) //题意为表明按数的顺序输入,比较i和j的大小 { n=i; i=j; j=n; } for(temp=i,max=0;temp<=j;temp++ //每次循环后要初始化max { n=temp; k=1; while(n!=1) { k++; if(n%2!=0) n=n*3+1; else n/=2; } if(max<k) max=k; } printf("%d\n",max); } return 0;
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