pOJ 2115

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35461		Accepted: 10372

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

来源：http://poj.org/problem?id=2115

程序与解析：

//poj2115
//一句话题意，问循环多少次后停止，如果循环此处超过2^k，则%2^k;
//假设循环x次(A+Cx)%2^k=B,得Cx+2^ky=B-A 
//(A+Cx)%2^k=B,即B= (A+Cx)-(A+Cx)/ 2^k*2^k
//令 (A+Cx)/ 2^k=y,则有 B= (A+Cx)-2^k*y,得B-A=Cx-2^k*y 
//令y=-y,则有 Cx+2^ky=B-A 
#include<iostream>
#include<cstdio>
using namespace std;
long long  A,B,C,k;
long long a,b,c,g,x,y;
void exgcd(long long a,long long b){
    if(b==0){//ax+0y=a,可以约定y=0,x只能为1 
        g=a;//全局变量 
        x=1;//全局变量
        y=0;//全局变量
        return ;
    }
    exgcd(b,a%b);
    long long  z=x;
    x=y;
    y=z-a/b*y;     
}
int main(){
    while (1){
        scanf("%d%d%d%d",&A,&B,&C,&k);
        if (A==0&&B==0&&C==0&&k==0) return 0;
        long long te=1;
        a=C;
        b=te<<k;//注意1<<31位是负数，因为默认1是int,最高位是符号位，此时可以long long(1)<<31，将1的存储空间强制转换为长整型。 
        c=B-A;
        exgcd(a,b);
        if(c%g!=0)printf("FOREVER
");
        else{
            x=x*c/g;//ax+by=c 与 ax+by=gcd(a,b)的关系 
            long long t=b/g;//符合ax+by的解x的公差是b/g. 
            //cout<<g<<" "<<t<<endl;
            x=(x%t+t)%t;//保证解x是正数。 
            printf("%lld
",x);//注意输出的是长整型 
        }
    }
    return 0;
}