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.
题解：
这个题的意思就是A需要加多少次C才能等于B
可以转化为： A+C*x≡B(mod k)
把模去掉： C*x+K*y=B-A
这就转化为扩展欧几里得了
不懂拓展欧几里得的小伙伴可以看我写的关于扩展欧几里得的博客
代码：
#include<iostream>
#include<stdio.h>
typedef long long ll;
int extgcd(ll a,ll b,ll &x,ll &y)
{
ll d=a;
if(b!=0)
{
d=extgcd(b,a%b,y,x);
y-=(a/b)*x;
}
else{
x=1; y=0;
}
return d;
}
int main()
{
long long a,b,c,k;
while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k)&&(a||b||c||k))
{
ll x,y;
ll n=(ll)1<<k; //求2的k次方
ll d=extgcd(c,n,x,y);
ll m=b-a;
if(m%d)
printf("FOREVER\n");
else
printf("%I64d\n",((x*m/d)%(n/d)+(n/d))%(n/d));
}
return 0;
}