POJ - 2115 - C Looooops （扩展欧几里得）

C Looooops

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

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

还是扩展欧几里得，这里注意要简化一下原来的式子

AC代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

LL gcd(LL a, LL b) {
	return b == 0 ? a : gcd(b, a % b);
}

void exgcd(LL a, LL b, LL& x, LL& y) {
	if(b == 0) {
		x = 1; y = 0;
	}
	else {
		exgcd(b, a % b, y, x);
		y -= x * (a / b);
	}
}

int main() {
	LL A, B, C, k;
	while(scanf("%I64d %I64d %I64d %I64d", &A, &B, &C, &k) != EOF) {
		if(A == 0 && B == 0 && C == 0 && k == 0) break;

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

		LL a = C;
		LL b = (1LL << k);
		LL c = gcd(a, b);
		LL d = B - A;

		if(d % c != 0) {
			printf("FOREVER\n");
			continue;
		}

		a /= c;//这里要进行简化，因为可能产生多余的次数
		b /= c;
		d /= c;

		LL p, q;
		exgcd(a, b, p, q);//这里求的是最简ax+by=gcd(a,b)的一组x,y的解 

		printf("%I64d\n", (p * (d / gcd(a, b)) % b + b) % b);

	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。