C Looooops（欧几里德+poj2115）

最新推荐文章于 2019-08-14 11:43:11 发布

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#C Looooops #poj2115 #扩展欧几里德 #欧几里德 #数论

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本文解析了一个特定形式的C语言循环，在给定初始值、终止条件及步长的情况下，探讨了循环体执行次数的计算方法。通过将问题转化为数学方程求解，利用扩展欧几里得算法找到可行解，并考虑到取模运算的特殊性。

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H - C Looooops

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2115

Appoint description:

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

题意：for(i=A;i!=B;i+=C){i%(2^k)};问你循环执行几次？

思路：先假设等式成立：(A+x*C)%(2^k)=B

变形(2^k)*y+B=A+C*x ==> C*x+(-(2^k)*y)=B-A;

ax+by=c

所以现在你知道怎么做了吧。哈哈！

转载请注明出处：http://www.cnblogs.com/yuyixingkong/

题目链接：http://poj.org/problem?id=2115

#include<stdio.h>
#define LL unsigned long long
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-=x*(a/b);
    }
}
int main()
{
    LL A,B,C,k;
    while(scanf("%llu%llu%llu%llu",&A,&B,&C,&k),(A+B+C+k))
    {
        LL a,b,c,d,x,y,dm;
        c=B-A;
        if(c==0){printf("0\n");continue;}
        a=C;
        b=(LL)1<<k;
        exgcd(a,b,d,x,y);
        if(c%d){ printf("FOREVER\n");continue;}
        dm=b/d;
        x=(((x*c/d)%dm)+dm)%dm;

        printf("%llu\n",x);
    }
    return 0;
}