poj2115--C Looooops(扩展gcd)

C Looooops

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 < 2^k) modulo 2^k.

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 < 2^k) 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

Source

题目大意初始值是a，每一次加c ，可以对得到的值取模2^k，在值为b时结束，问最少会执行多少次，如过一直不会停FOREVER

由题中大意的到等式，假设一定会停，那么在循环x次后会等于b   也就是  （a + x*c）% （2^k）= b 。很明显的扩展gcd的题目，得到等式， a + c*x - b = y*(2^k) 也就是

求解 c*x - (2^k)*y = b-a ,问有没有整数解。

对于a*x1 + b*y1 = c ，用扩展gcd，求解 a*x1 + b*y1 = gad(a,b) = gcd(b,a%b)  = a*y2 +b* (x2-a/b*y2)得到  x1 = y2 , y1 = ( x2 +a/b*y2 ) , 利用这个等式，可以推到b = 0 时，那是x = 1 , y = 0 ,最大公约数为a，在逐层返回，最终得到 a*x1 + b*y1 = gad(a,b) 的解，如果c%( gcd(a,b) ) == 0 那么a*x1 + b*y1 = c有整数解，令d =
gcd(a,b)，d = b / d.最小的正整数解围 x = ( x%d +ｄ )%d.

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL __int64
using namespace std;
void gcd(LL a,LL b,LL &d,LL &x,LL &y)
{
    if(b == 0)
    {
        d = a ;
        x = 1 ;
        y = 0 ;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y = -y ; x = -x ;
        y += a/b*x ;
    }
    return ;
}
int main()
{
    LL k , a , b , c , d , x , y ;
    while(scanf("%I64d %I64d %I64d %I64d", &a, &b, &c, &k)!=EOF)
    {
        if(a == 0 && b == 0 && c == 0 && k == 0)
            break;
        k = ( (LL)1 )<<k ;
        gcd(c,k,d,x,y);
        if( (b-a)%d )
            printf("FOREVER\n");
        else
        {
            x = (b-a)/d*x ;
            k = k/d ;
            x = (x%k+k)%k ;
            printf("%I64d\n", x);
        }
    }
    return 0;
}

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