poj2115——C Looooops（数论，解模线性方程）

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

基本算是《算法导论（第三版）》上的例题了，欧几里得扩展算法在p549，模线性方程的解法在p555。本体与书上不同的地方在于，书上是解ax=b(mod n)的所有解，而题目只需要最小的正整数输出，所以还需要用x=(x%(n/d)+n/d)%(n/d)来得出需要的那个解。

还有几个小技巧，2^k可以直接用1<

#include <iostream>
#include <cstdio>
using namespace std;
long long gcd(long long a,long long b,long long &x,long long &y)
{
if(b==0)
{
x=1;
y=0;
return a;
}
else
{
long long d=gcd(b,a%b,x,y);
long long xx=x;
x=y;
y=xx-a/b*y;
return d;
}
}
int main()
{
long long A,B,C,k,a,b,n,d,x,y;
while(scanf("%I64d%I64d%I64d%I64d",&A,&B,&C,&k))
{
if(A==0&&C==0&&C==0&&k==0)
break;
a=C;
b=B-A;
n=1LL<<k;
d=gcd(a,n,x,y);
if(b%d==0)
{
x=x*(b/d)%n;
x=(x%(n/d)+n/d)%(n/d);
printf("%I64d\n",x);
}
else
printf("FOREVER\n");
}
return 0;
}