C Looooops（扩展欧几里得）题解

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

思路：

和之前做的差不多：青蛙

发现模线性方程不会转化orz，贴一下：模线性方程

模线性方程转化：当 a*x≡b mod m时，可以转化为 a*x+m*y=b求解x，y

那么我们先来列出原题模线性方程：C*t+A ≡ B mod(2^k)  ==>   C*t ≡ B-A  mod(2^k) ==>  C*t + (2^k)*y = B-A

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
const int N=2000000000;
const int MAX=2100000000;
const int MOD=1000;
using namespace std;
ll ex_gcd(ll a,ll b,ll &x,ll &y){
	ll d,t;
	if(b==0){
		x=1;
		y=0;
		return a;
	}
	d=ex_gcd(b,a%b,x,y);
	t=x-a/b*y;
	x=y;
	y=t;
	return d;
}
int main(){
	ll a,b,c,k,A,B,C,d,x,y;
	while(~scanf("%lld%lld%lld%lld",&A,&B,&C,&k) && A+B+C+k){
		a=C;
		b=(ll)1<<k;
		c=B-A;
		d=ex_gcd(a,b,x,y);
		if(c%d!=0){
			printf("FOREVER\n");
		}
		else{
			x=x*c/d;
			ll k=b/d;
			x=(x%k+k)%k;
			printf("%lld\n",x);
		}
	}
	return 0;
}