poj 2115 C Looooops(解线性同余方程)

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19017		Accepted: 5003

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

CTU Open 2004
题目分析：
解线性同余方程：
首先这道题的方程很容易得到
a+cy == b ( mod (1<<k) )
那么化成一般形式就是
a+cx +(1<<k)y = b - a
然后求取d=gcd(c,(1<<k) )，然后如果d|(b-a)不成立，那么无解
如果成立,被d除，之后，利用欧几里得求解之后得到一个解，
然后*d得到一个正解，然后模上(1<<k)/d，获得最小整数解，因为解的集合为x=x1+(1<<k)/d*i ( ０＜= 0 < d )

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

LL a,b,c,k;

void exgcd ( LL a , LL b , LL& d , LL& x , LL&y )
{
    if ( !b )
        d = a , x = 1 , y = 0;
    else
        exgcd ( b , a%b , d , y , x ) , y -= x*(a/b);
}

LL f ( LL  a , LL b , LL m )
{
    LL d,x,y;
    exgcd ( a, m , d , x , y );
    if ( b % d )
        return -1;
    x = (x*(b/d)%m + m )%m;
    x = (x+(m/d))%(m/d);
    return x;
}

int main ( )
{
    while ( ~scanf ( "%lld%lld%lld%lld" , &a , &b , &c , &k ) )
    {
        if ( !a && !b && !c && !k ) break;
        LL temp = f ( c , b-a , (LL)1<<k );
        if ( temp == -1 )
            puts ( "FOREVER" );
        else printf ( "%lld\n" , temp );
    }
}