Uva 3n+1 问题

3n+1 问题:

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 whoseclassification 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 2Given the input 22, the following sequence of numbers will be printed22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral inputvalue. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It hasbeen verified, however, for all integers n such that 0 < n < 1, 000, 000 (and, in fact, for many morenumbers than this.)Given an input n, it is possible to determine the number of numbers printed before and includingthe 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cyclelength of 22 is 16.For any two numbers i and j you are to determine the maximum cycle length over all numbersbetween and including both i and j.InputThe input will consist of a series of pairs of integers i and j, one pair of integers per line. All integerswill be less than 10,000 and greater than 0.You should process all pairs of integers and for each pair determine the maximum cycle length overall integers between and including i and j.You can assume that no operation overflows a 32-bit integer.OutputFor each pair of input integers i and j you should output i, j, and the maximum cycle length forintegers between and including i and j. These three numbers should be separated by at least one spacewith all three numbers on one line and with one line of output for each line of input. The integers iand j must appear in the output in the same order in which they appeared in the input and should befollowed by the maximum cycle length (on the same line).Sample Input1 10100 200201 210900 1000Sample Output1 10 20100 200 125201 210 89900 1000 174

#include<stdio.h>int main (){    int i,j,n,m,k,b,c,f,t;    while(scanf("%d%d",&i,&j)!=EOF)    {         c=0;         f=0;         if(i>j)         {             t=i;             i=j;             j=t;             f=1;         }        for(k=i; k<=j; k++)        {            n=1;            b=k;            while(b!=1)            {                if(b%2!=0)                {                    b=b*3+1;                }                else                    b=b/2;                n++;            }           if(c<n)            c=n;        }        if(f==0)        printf("%d %d %d\n",i,j,c);        else            printf("%d %d %d\n",j,i,c);    }    return 0;}

—-2017/7/21

总结：由于习惯性的思维认为i < j,导致错误。