POJ 2115-C Looooops-擴充套件歐幾里德演算法

kewlgrl發表於2016-03-09
OOP套件演算法
C Looooops
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 21526   Accepted: 5868

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

Source

CTU Open 2004


知識點：
詳細題解
歐幾里德演算法
擴充套件歐幾里德演算法
求解模線性方程


#include<iostream>
using namespace std;

//d=ax+by,其中最大公約數d=gcd(a,b)，x、y為方程係數，返回值為d、x、y
__int64 EXTENDED_EUCLID(__int64 a,__int64 b,__int64& x,__int64& y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;  //d=a，x=1,y=0,此時等式d=ax+by成立
	}
	__int64 d=EXTENDED_EUCLID(b,a%b,x,y);
	__int64 xt=x;
	x=y;
	y=xt-a/b*y;  //係數x、y的取值是為滿足等式d=ax+by
	return d;
}

int main(void)
{
	__int64 A,B,C,k;
	while(scanf("%I64d %I64d %I64d %I64d",&A,&B,&C,&k))
	{
		if(!A && !B && !C && !k)
			break;

		__int64 a=C;
		__int64 b=B-A;
		__int64 n=(__int64)1<<k;  //2^k
		__int64 x,y;
		__int64 d=EXTENDED_EUCLID(a,n,x,y);  //求a,n的最大公約數d=gcd(a,n)和方程d=ax+by的係數x、y

		if(b%d!=0)  //方程 ax=b(mod n) 無解
			cout<<"FOREVER"<<endl;
		else
		{
			x=(x*(b/d))%n;  //方程ax=b(mod n)的最小解
			x=(x%(n/d)+n/d)%(n/d);  //方程ax=b(mod n)的最整數小解
			printf("%I64d\n",x);
		}
	}
	return 0;
}


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