POJ 2115 C Looooops（扩展欧几里得）

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

第一次做扩欧的题，也是因为这个题才学习了扩展欧几里得，感觉数学的世界简直是太神奇也太复杂了，搞了一晚上都搞晕乎了，扩展欧几里得就不多说了，我也是刚学会也不会讲太多，这道题的意思是给出初始值a，循环终止值b和增量c，问在k位存储体系中循环多少次可以终止，也就是转化成了求出cx+2^ky=(b-a)的一组可行解，根据扩展欧几里得，如果(b-a)不可以被gcd(c,2^k)整除，那么该方程无解，输出“FOREVER”，可以整除则至少存在一组可行解，扩展欧几里得算法便求出了其中一解。

根据这题题意，对于循环次数x，x=[(B-A+2^k)%2^k] /C

即 Cx=(B-A)(mod 2^k)  此方程为模线性方程，本题就是求x的值。
即求模线性方程 ax=b( mod n)的解。 其中a=C，b=B-A，n=2^K.
该方程有解的充要条件为b% gcd(a,n)==0
令d=gcd(a,n)有该方程的 最小整数解为 x0 = X mod (n/d)
X为任意解,x0为最小整数解

这题唯一需要注意的就是k=32时int会溢出，所以要用long long 才可以。

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <map>
#include <set>
#include <bitset>
#include <ctime>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <ctime>

#define STD_REOPEN() freopen("../in.in","r",stdin)
#define STREAM_REOPEN fstream cin("../in.in")
#define INF 0x3f3f3f3f
#define _INF 63
#define eps 1e-4
#define MAX_V 100010
#define MAX_P 2010
#define MAX_E 4001000
#define MAX 32000
#define MOD_P 3221225473

using namespace std;

long long extendGCD(long long a,long long b,long long &x,long long &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return a;
	}
	long long ans=extendGCD(b,a%b,x,y);
	long long t=x;
	x=y;
	y=t-a/b*y;
	return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    long long a,b,c,k,x,y;
    while(cin>>a>>b>>c>>k,a||b||c||k)
    {
		long long d=1;
		d<<=k;
		//cout<<d<<endl;
		long long e=extendGCD(c,d,x,y);
		if((b-a)%e)
			cout<<"FOREVER"<<endl;
		else
		{
			long long t=d/e;
			x=(x*(b-a)/e%t+t)%t;
			cout<<x<<endl;
		}
    }

    return 0;
}