2115 C Looooops 拓展欧几里得

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本文探讨了一种特定形式的C语言for循环，并提供了一个程序来计算该循环在不同参数设置下执行的次数。通过数学方法解决循环终止条件，确定循环是否能够结束及执行的具体次数。

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8812		Accepted: 1969

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

Sample Output

0
2
32766
FOREVER

#include<iostream>
#include<cstdio>
using namespace std;
__int64 exgcd(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    __int64 r=exgcd(b,a%b,x,y);
    __int64 t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
int main()
{
    __int64 a,b,c,k;
    while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k)==4)
    {
        if(a==0&&b==0&&c==0&&k==0) break;
        __int64 n=c,m=(__int64)1<<k,v=b-a,x,y;//m=(__int64)1<<k  不能写成(__int64)(1<<k),否则越界
        __int64 r=exgcd(n,m,x,y);//nx = v(mod m)
        if(v%r)
        {
            printf("FOREVER/n");
            continue;
        }
        x=x*(v/r)%m+m;
        __int64 g=m/r;
        x=((x%g)+g)%g;
        printf("%I64d/n",x);
    }
    return 0;
}