poj2115——拓展欧几里德求模线性同余方程的最小正整数解

poj2115——拓展欧几里德求模线性同余方程的最小正整数解

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18926		Accepted: 4973

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：for(x=A;x!=B;x+=C,x%=1LL<<k) 求解结束时循环的次数
思路：显然是求解方程(A+Cx)%n=B，(n=1<<k)，变形得Cx%n=(B-A)%n，即转为模线性同余方程 Cx=(B-A)(mod n)，扩展gcd解之即可

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<algorithm>



using namespace std;



const int maxn=1000100;

const int INF=(1<<28);

typedef long long ll;



ll A,B,C,k;

ll x,y;



ll exgcd(ll a,ll b, ll &x, ll &y)

{

    if(b==0){

        x=1;y=0;

        return a;

    }

    ll r=exgcd(b,a%b,x,y);

    ll t=y;

    y=x-a/b*y;

    x=t;

    return r;

}



bool mod_linear_equation(ll a,ll b,ll n)

{

    ll d=exgcd(a,n,x,y);

    if(b%d) return false;

    x=x*(b/d)%n;

    x=(x%(n/d)+n/d)%(n/d);

    return true;

}



int main()

{

    while(cin>>A>>B>>C>>k,A||B||C||k){

        if(mod_linear_equation(C,B-A,1LL<<k)) cout<<x<<endl;

        else cout<<"FOREVER"<<endl;

    }

    return 0;

}

View Code