poj2115 C Looooops

最新推荐文章于 2019-02-21 10:19:24 发布

LL_Sagiri

最新推荐文章于 2019-02-21 10:19:24 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，通过数学分析给出了解决方案，并提供了完整的代码实现。适用于k位无符号整数类型的循环，探讨了循环终止条件及执行次数。

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C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30277		Accepted: 8748

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

题解：

A+nC-B=m2^k

nC-m2^k=B-A

exgcd求即可。

代码：

#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;  
}