poj 2115 二元一次不定方程

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14765		Accepted: 3719

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

题目大意：计算循环的次数，有a+ct=b(mod 2^k),变换下得到二元一次不定方程：ct+p*2^k=b-a;
用B=1<<k Wrong answer
正确的 B=pow(2,k+0.0);
AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

__int64 a,b,c,A,B,C,x,y,d,t,k;

__int64 Extended_Euclid(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
    __int64 d,t;
    if(b==0)
    {
        x=1;y=0;
        return a;
    }
    d=Extended_Euclid(b,a%b,x,y);
    t=x;
    x=y;
    y=t-a/b*y;
    return d;
}

int main()
{
    while(scanf("%I64d %I64d %I64d %d",&a,&b,&c,&k),a+b+c+k)
    {
        if (a==b) {cout<<0<<endl; continue;}
        else if (c==0) {cout<<"FOREVER"<<endl; continue;}
        A=c;
        C=b-a;
        B=pow(2,k+0.0);    
        d=Extended_Euclid(A,B,x,y);
        if(C%d)
            printf("FOREVER\n");
        else 
        {
            t=B/d;
            x=x*C/d;
            x=(x%t+t)%t;
            printf("%I64d\n",x);
        }
    }
    return 0;
}