3 n+1デジタル処理

HDU - 1039
タイトルの説明
Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22: 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000. For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.
 
入力
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.
 
しゅつりょく
For each pair of input integers i and j, output i, j in the same order in which they appeared in the input and then the maximum cycle length for integers between and including i and j. These three numbers should be separated by one space, with all three numbers on one line and with one line of output for each line of input.
 
サンプル入力
1 10
100 200
201 210
900 1000
 
サンプル出力
1 10 20
100 200 125
201 210 89
900 1000 174
 
に言及
nとmを入力し、nからmまでの任意の数を判断し、偶数は2で割って奇数は3に1を加えて結果が1になるまで、入力したm,nを元の順序で、処理の最大回数も、3つの数の間にスペースを含んで出力し、各行の出力が1行を占める.
 
解体構想:
forループを用いて処理を行い,各処理の回数を配列に格納し,比較により最も多い回数を求める.
エラーは1回、元の入力順序でn,mを出力すべきで、私は小さいから大きいまで出力するだけで、だからエラー;
交換した値があるかどうかを変数で観察し,交換した場合は逆シーケンスで出力する.
 
コード編:
 

#include
int main()
{
    int n,max,b,i,j,t,m,y,x,a[10000];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        b=0;
        if(n>m)//    ，        ， b        ；
        {
            t=n;n=m;m=t;
            b++;
        }
        i=0,max=0,j=0;
        for(x=n;x<=m;x++)
        {
            i=1,y=x;
            while(y!=1)// 1    ；
            {
                if(y%2==0)
                    y/=2;
                else
                    y=y*3+1;
                i++;//       ；
            }
            a[j]=i;//         ；
            j++;
        }
        for(i=0;imax)
                max=a[i];
        }
        if(b==1)//    ，     ；
        {
            printf("%d %d %d
",m,n,max);
        }else
        printf("%d %d %d
",n,m,max);
    }
    return 0;
}