poj 2115 C Looooops His love for her is enshrined forever in his heart.扩展欧几里得-优快云博客

本文探讨了一种特殊的C语言for循环，在给定的参数下计算循环体内语句的执行次数。通过将问题转化为求解特定数学方程的形式，并利用扩展欧几里得算法来解决。特别关注了位运算的使用以提高效率。

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15519		Accepted: 3962

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

Source

CTU Open 2004

这个题目的意思是说C语言的for循环，我们问是否使v=a；v=v+c,在循环中使v=b,如果不能的话就是forever了，同时还要使其%（2^k),才可以，也就是说为（a+x*c)%2^k=b,

那么就是c*x=(b-a)%2^k,其实也就可以换成一个式子x*A+B*y=C，这样就成功比对了欧几里得定理的公式，C=b-a,A=c,B=2^k,这样我们在做的时候直接套用欧几里得定理就是了，就这么简单！！！！，另外对于2^k，的使用，我们要用位运算，用pow编译器不会报错，但是提交就会错了！！！！

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
long long exgcd(long long a,long long b,long long &x,long long &y)
{
    long long d,t;
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    d=exgcd(b,a%b,x,y);
    t=x-a/b*y;
    x=y;
    y=t;
    return d;

}
int main()
{
    long long a,b,x,y,c,a1,a2,a3,k;
    int i,j;
    while(scanf("%lld%lld%lld%lld",&a1,&a2,&a3,&k)!=EOF)
    {
        if((a1==0)&&(a2==0)&&(a3==0)&&(k==0))
        break;
        a=a3;
        //b=pow(2,k);
        b=1LL<<k;
        c=a2-a1;
        long long r=exgcd(a,b,x,y);
        if(c%r)
        {
            printf("FOREVER\n");

        }
        else
        {
            long long  s=b/r;
            x=x*(c/r);
            x=(x%s+s)%s;
            cout<<x<<endl;
        }

    }
    return 0;
}