POJ 2115：C Looooops

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19536		Accepted: 5204

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

题意是问在

for (variable = A; variable != B; variable += C)

这种情况下，循环多少次。

其中所有的数要mod 2的k次方。所以方程就是(A+C*x)%(2^k)=B，变换一下就是-C*x+(2^k)*y=A-B。解这个方程的最小正数x即可。

又是扩展欧几里德。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

long long yue;

void ex_gcd(long long a,long long b, long long &xx,long long &yy)
{
	if(b==0)
	{
		xx=1;
		yy=0;
		yue=a;
	}
	else
	{
		ex_gcd(b,a%b,xx,yy);

		long long t=xx;
		xx=yy;
		yy=t-(a/b)*yy;
	}
}


int main()
{
	long long A,B,C,k,k2,xx,yy;

	while(scanf_s("%lld%lld%lld%lld",&A,&B,&C,&k))
	{
		if(!A&&!B&&!C&&!k)
			break;

		k2=(1LL<<k);
		ex_gcd(-C,k2,xx,yy);

		if((A-B)%yue)
		{
			cout<<"FOREVER"<<endl;
		}
		else
		{
			xx=xx*((A-B)/yue);
			long long r=k2/yue;
			if(r<0)
				xx=(xx%r-r)%r;
			else
				xx=(xx%r+r)%r;
			printf("%lld\n",xx);
		}
	}
	return 0;
}

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