POJ 2115 C Looooops

                                                            C Looooops

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

考察扩展欧几里得算法：

#include <stdio.h>
#include <math.h>

long long extended_euclid(long long a, long long b, long long &x, long long &y )
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;     //d=a，x=1,y=0,此时等式d=ax+by成立
	}
	int d=extended_euclid(b, a%b, x, y );
	int xt=x;
	x=y;
	y=xt-a/b*y;
	return d;
}


int main()
{
	int A, B, C, k;
	long long a, b, c;
	long long n;

	while(scanf("%d %d %d %d", &A, &B, &C, &k)!=EOF)
	{
	    if(A==0 && B==0 && C==0 && k==0)
	        break;
		n=1;
		for(int i=1; i<=k; i++)
		{
			n=n*2;
		}
		a=C;
		b=B-A;
		long long x, y;
        long long d=extended_euclid (a, n, x, y );

		if(b%d!=0) //无解
		{
			printf("FOREVER\n");
			continue ;
		}
        x=(x*(b/d))%n;
        x=(x%(n/d)+n/d)%(n/d);

		printf("%lld\n", x);
	}
	return 0;
}

 别人写的的代码+解析：

 

  

#include<iostream>  
using namespace std;  
  
//d=ax+by,其中最大公约数d=gcd(a,n)，x、y为方程系数，返回值为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,此时等式d=ax+by成立  
    }  
    __int64 d=EXTENDED_EUCLID(b,a%b,x,y);  
    __int64 xt=x;  
    x=y;  
    y=xt-a/b*y;  //系数x、y的取值是为满足等式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的最大公约数d=gcd(a,n)和方程d=ax+by的系数x、y  
  
        if(b%d!=0)  //方程 ax=b(mod n) 无解  
            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)的最整数小解  
            printf("%I64d\n",x);  
        }  
    }  
    return 0;  
}