POJ 2115 C Looooops 扩展欧几里德算法

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

题目大意：给你一个循环, for (int i=a, i!=b; i+=c), 并且告诉你计算机是多少位的操作系统，意思就是i到达多少的时候取模，比如65535，当i 大于 65535时，在从1开始算。这也是一道简单的拓展欧几里德模板题。
对于输入的数据可以列式子：A+KC mod 2k = B 转换后得到Cx + 2^ky = B-A, 直接套模板就可以了

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define inf 1000010
using namespace std;
ll exgcd(ll a,ll b,ll &x,ll &y){
    ll d=a;
    if(b!=0){
        d=exgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }else{
        x=1;y=0;
    }
    return d;
}
int main(){
    ll A,B,C,k,a,b,x,y,c,g;
    while(scanf("%lld %lld %lld %lld",&A,&B,&C,&k)!=EOF&&A+B+C+k){
        a=C;b=1LL<<k;c=B-A;//不能写成1<<k;
        g=exgcd(a,b,x,y);
        if(c%g){
            printf("FOREVER\n");
        }else{
            b/=g;
            c/=g;
            x=((x*c)%b+b)%b;
            printf("%lld\n",x);
        }
    }
    return 0;
}