poj-2115-exgcd

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32062		Accepted: 9337

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

Source

CTU Open 2004

 

　　在模2^k的意义下，A每次加C，最少加多少次能达到B，输出这个次数。

　　易得 A+i*C= B  (mod 2^k)   -->   i*C+j*(2^k) = B - A ,问题转化为求这个方程成立的最小的i值，直接拓欧，坑点是K最大32，不强制LL

的话会爆。

　　

#include<iostream>
#include<cstdio>
using namespace std;
#define LL long long 
#define mp make_pair
#define pb push_back
#define inf 0x3f3f3f3f
void exgcd(LL a,LL b,LL &d,LL &x,LL &y){
    if(!b){d=a;x=1;y=0;}
    else{exgcd(b,a%b,d,y,x);y-=(a/b)*x;}
}
int main(){
    LL A,B,C,K;
    while(scanf("%lld%lld%lld%lld",&A,&B,&C,&K)!=EOF){
        if(A==0&&B==0&&C==0&&K==0) break;
        LL x,y,d;    
        LL M=1;
        M<<=K;
        exgcd(C,M,d,x,y);
        if((B-A)%d!=0){
            puts("FOREVER");
        }
        else{
            M/=d;
            printf("%lld\n",((x*(B-A)/d)%M+M)%M);
        }
    }
    return 0;
}