POJ 2115 C Looooops

最新推荐文章于 2019-07-15 06:56:00 发布

原创最新推荐文章于 2019-07-15 06:56:00 发布 · 777 阅读

0 ·

CC 4.0 BY-SA版权

ACM/数论专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种计算特定C语言循环执行次数的方法，利用同余方程和扩展欧几里得算法来解决循环终止条件下的执行次数问题。适用于k位无符号整数计算环境。

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14193		Accepted: 3547

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

题目链接

知识点：同余方程 + 扩展欧几里得算法

#include<cstdio>

using namespace std;

long long num[33];

#define LL long long
#define lld	lld

LL ext_gcd(LL a,LL b,LL &x,LL &y)
{
	LL tmp,ret;
	if(!b)
	{
		x=1;	y=0;	return a;
	}
	ret=ext_gcd(b,a%b,x,y);

	tmp=x;	x=y;	y=tmp-a/b*y;
	return ret;
}

int main()
{
	LL A,B,C,k,tmp;
	LL i,j,x,y,g;

	num[1]=2;
	for(i=2;i<=32;i++)
		num[i]=num[i-1]*2;

	while(scanf("%lld%lld%lld%lld",&A,&B,&C,&k)==4 && A||B||C||k)
	{
		g=ext_gcd(C,num[k],x,y);
		if((A-B)%g==0)
		{
			tmp=x*(B-A)/g;
			tmp=(tmp%(num[k]/g)+num[k]/g)%(num[k]/g);
			printf("%lld\n",tmp);
		}
		else
			printf("FOREVER\n");
	}

	return 0;
}