POJ 2115 ( 中国剩余定理 )

                                                    C Looooops
Time Limit: 1000MS
 
Memory Limit: 65536K
Total
 Submissions: 17625
 
Accepted: 4563
Description
A Compiler Mystery: We are given a C-language style for loop of type 
for (variable = A; variable != B; variable += C)

  statement;
I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2k) modulo 2k. 

Input
The input consists of several instances. Each instance is described by a single line with four integers A, B, C, k separated by a single space. The integer k (1 <= k <= 32) is the number of bits of the control variable of the loop and A, B, C (0 <= A, B, C < 2k) are the parameters of the loop. 

The input is finished by a line containing four zeros. 
Output
The output consists of several lines corresponding to the instances on the input. The i-th line contains either the number of executions of the statement in the i-th instance (a single integer number) or the word FOREVER if the loop does not terminate. 
Sample Input
3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0
Sample Output
0
2
32766
FOREVER

题意：变量初始为A，然后每次加C，问需要几步才能等于 B。不过每次都要对 2的k次方 取模。 即 A+ xC= B( mod 1<<k) 

整理得：Cx  = B-A （ mod 1<<k）----->Cx + (1<<k) y = B-A----->aX + bY = c   (  其中a = C,   b = 1<<k, c = B-A)。

  然后再利用拓展欧几里得求最小正整数x即可。

#include<stdio.h>
#include<math.h>
#define LL long long
LL a, b, c ,k, x, y;

int ext_gcd(LL a, LL b, LL &x, LL &y)
{
    if(b==0) {
        x=1;
        y=0;
      return a;
    }
    LL u = ext_gcd(b, a%b, x ,y);
    LL t = x;
       x = y;
       y = t-a/b*y;
    return u;
}

int main()
{
    while(~scanf("%lld%lld%lld%lld", &a, &b, &c, &k))
    {
        if(a==0 && b==0 && c==0 && k==0)
              break;
        LL C = b-a, A = c, B = (LL)1<<k;
        LL d = ext_gcd(A, B, x, y);
        if(C%d!=0)
        {
            printf("FOREVER
");
        }
        else
        {
            x = x*C/d;
            B = B/d;
            x = (x%B+B)%B;
            printf("%lld
", x);
        }
    }
    return 0;
}