POJ C Looooops 2115 (擴展歐幾里得)

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23143		Accepted: 6357

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

Source

CTU Open 2004

題意：給出a,b,c,k,算(a+x*c)%2^k是否等於b

分析：擴展歐幾里得最小整數解　可以把上面的方程轉成　a*x + 2^k*y = b

如果沒有整數解，則永遠到不了，套模板，這裏有模板點擊打開鏈接

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
using namespace std;
long long x,y;
long long e_gcd(long long a,long long b,long long &x,long long &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    long long ans=e_gcd(b,a%b,x,y);
    long long temp;
    temp=x;
    x=y;
    y=temp-a/b*y;
    return ans;
}
int main()
{
    long long a,b,c,d,n,m,k,l;
    while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&k))
    {
        if(a==0 && b==0 && c==0 && k==0)
            break;
        n=c;
        m=1LL<<k;
        l=b-a;
        d=e_gcd(n,m,x,y);
        if(l%d!=0)
            printf("FOREVER\n");
        else
        {
            x=x*(l/d);
            m=m/d;
            printf("%lld\n",((x%m)+m)%m);
        }
    }
    return 0;
}