POJ2115 C Looooops

Posted 2020-10-07 noble_

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

--------------------------------------------------------------------------
题意：在一个k位的机器里（大于2^k就回到0），求循环何时终止。
分析：似乎是裸的扩欧。方程：c*x + 2^k*y = b-a 。

 1 #include <cstdio>
 2 typedef long long LL;
 3 LL exgcd(LL a,LL b,LL &x,LL &y)
 4 {
 5     int d;
 6     if(b==0)
 7     {
 8         x=1;y=0;return a;
 9     }
10     else
11     {
12         d=exgcd(b,a%b,y,x);y-=x*(a/b);
13     }
14     return d;
15 }
16 int main()
17 {
18     LL a,b,c,k;
19     while(scanf("%lld%lld%lld%lld",&a,&b,&c,&k)&&(a||b||c||k))
20     {
21         
22         LL i=b-a,x=0,y=0,d=0,p=1LL<<k;//不加LL会爆 
23         //方程：c*x + 2^k*y = b-a 
24         d=exgcd(c,p,x,y);
25         if(i%d!=0) 
26         {
27             printf("FOREVER\n");
28             continue;
29         }
30         p/=d;
31         x%=p;
32         x*=(i/d)%p;//把倍数乘上
33         x=(x%p+p)%p;
34         printf("%lld\n",x);
35     }
36     return 0;
37 }