HDU 4229 Vive la Difference!

Take any four positive integers: a, b, c, d. Form four more, like this:
|a-b| |b-c| |c-d| |d-a|
That is, take the absolute value of the differences of a with b, b with c, c with d, and d with a. (Note that a zero could crop up, but they’ll all still be non-negative.) Then, do it again with these four new numbers. And then again. And again. Eventually, all four integers will be the same. For example, start with 1,3,5,9:
1 3 5 9
2 2 4 8 (1)
0 2 4 6 (2)
2 2 2 6 (3)
0 0 4 4 (4)
0 4 0 4 (5)
4 4 4 4 (6)
In this case, the sequence converged in 6 steps. It turns out that in all cases, the sequence converges very quickly. In fact, it can be shown that if all four integers are less than 2^n, then it will take no more than 3*n steps to converge!
Given a, b, c and d, figure out just how quickly the sequence converges.

 

Input

There will be several test cases in the input. Each test case consists of four positive integers on a single line (1 ≤ a,b,c,d ≤ 2,000,000,000), with single spaces for separation. The input will end with a line with four 0s.

 

Output

For each test case, output a single integer on its own line, indicating the number of steps until convergence. Output no extra spaces, and do not separate answers with blank lines.

 

Sample Input

1 3 5 94 3 2 11 1 1 10 0 0 0 

 

Sample Output

640 

#include <stdio.h>int pd(long long a,long long b,long long c,long long d){    if(a==b&&a==c&&a==d)        return 1;    return 0;}long long f(long long n){    return n>0?n:-n;}int main(){    long long a,b,c,d,e;    int i;    while(~scanf("%lld %lld %lld %lld",&a,&b,&c,&d),a||b||c||d)    {        for(i=0;;i++)        {            if(pd(a,b,c,d))                break;            e=a;            a=f(a-b),b=f(b-c),c=f(c-d),d=f(d-e);        }        printf("%d\n",i);    }    return 0;}