POJ 2115-C Looooops(扩展欧几里德)

最新推荐文章于 2019-08-14 11:43:11 发布

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本文介绍了一种计算特定C语言循环执行次数的方法，通过将循环条件转化为数学方程，利用扩展欧几里德算法求解未知数，进而得出循环终止时的执行次数。

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17981		Accepted: 4694

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

还是扩展欧几里德。。。

题意 ：有一条for循环语句 (int i=A;i!=B;i=(i+C)%2^k) sta; 给定A,B,C,k;问循环什么时候终止；

设循环x次终止， 则可得 (A+x*C)%2^k=B;即 A+x*C-B=y*2^k;(y=0,1,2...)

整理  C*x-2^k*y=B-A;令a=C;b=-2^k;c=B-A;

得  ax+by=c;

上式有解即意味着循环可以终止，x的最小整数解即为循环次数。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <vector>
#include <string>
#include <cmath>
#include <map>
#include <queue>
using namespace std;
#define LL long long
LL gcd(LL a,LL b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
void exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return ;
    }
    exgcd(b,a%b,y,x);
    y-=a/b*x;
}
int main()
{
    LL A,B,C,k,x,y;
    while(cin>>A>>B>>C>>k){
    if(!A&&!B&&!C&&!k)break;
    LL a=C;
    LL b=-pow(2LL,k);
    LL c=B-A;
    LL r=gcd(a,b);
    if(c%r)
    {
        puts("FOREVER");
        continue;
    }
    a/=r;b/=r;c/=r;
    exgcd(a,b,x,y);
    if(b<0)b=-b;
    x*=c;
    cout<<(x%b+b)%b<<endl;
  }
    return 0;
}