[poj 2115] C Looooops 扩展欧几里得

最新推荐文章于 2020-05-17 17:19:39 发布

原创最新推荐文章于 2020-05-17 17:19:39 发布 · 841 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#poj

poj 题目记录同时被 2 个专栏收录

70 篇文章

订阅专栏

数学(数论相关)

10 篇文章

订阅专栏

本文探讨了特定C语言风格for循环在k位无符号整型系统中的执行次数问题。通过数学推导，利用扩展欧几里德算法解决循环终止条件下的次数计算，以及判断是否进入死循环。

C Looooops
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 21706 Accepted: 5927

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 < 2k) modulo 2k.

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 < 2k) 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

题目链接：
http://poj.org/problem?id=2115

题意：对于C的for(i=A ; i!=B ;i +=C)循环语句，问在k位存储系统中循环几次才会结束。

若在有限次内结束，则输出循环次数。

否则输出死循环。

思路：
———————-(A+C*x)=B(mod n)———————;
即 Cx=(B-A)(mod 2^k)
=C
b=B-A
n=2^k
–》ax=b (mod n)
！！所以方程 a*x+b*y=n；我们可以先用扩展欧几里德算法求出一组x0，y0。也就是a*x0+b*y0=（a，b）；然后两边同时除以（a，b），再乘以n。这样就得到了方程a*x0*n/（a，b）+b*y0*n/（a，b）=n；我们也就找到了方程的一个解。！！
ax-ny=b 有解的充要条件为 gcd(a,n) | b ，即 b% gcd(a,n)==0
有该方程的最小整数解为 x = e (mod n/d)

其中e = [x0 mod(n/d) + n/d] mod (n/d) ，x0为方程的最小解

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
long long  A,B,C,k;
long long a,b;
long long x,y;

long long  ex_gcd(long long  a,long long  b,long long & x,long long  &y)
{

    if(b==0) 
    {
        x=1;
        y=0;
        return a;
    }
    long long d= ex_gcd(b,a%b,x,y);
    int t=x;
    x=y;
    y=t-a/b*y;
    return d;
}
int main()
{   
    while(scanf("%lld %lld %lld %lld",&A,&B,&C,&k))
    {
        if(!A&&!B&&!C&&!k) return 0;
        a=C;
        b=B-A;
        long long n=(long long )1<<k;
        long long d;
        d=ex_gcd(a,n,x,y);

        if(b==0) 
                 {
                        printf("0\n");
                        continue;
        }

        if(b%d==0) 
        {
             x=(x*(b/d))%n; 
            x=(x%(n/d)+n/d)%(n/d);
            printf("%lld\n",x);
        }
        else  printf("FOREVER\n");

    }
}