[POJ 2115] C Looooops【exgcd】

  • 2018-02-02
  • 0
  • 0

Problem:

Time Limit: 1000MS Memory Limit: 65536K

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

Solution:

简单数论题。

不难发现,设执行次数为 x,则

  • A + Cx ≡ B (mod 2k)
  • 可化为 Cx + 2ky = B - A

而 exgcd 正好可以解形如 ax + by = gcd(a, b) = d 的同余方程。

所以考虑放大或缩小的倍数,当且仅当 gcd(C, 2k) | (B - A) 时方程有解。

只要 exgcd 一下,然后将倍数还原即可。

Code: O(Tk) [684K, 0MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

ll A, B, C, ans;
int k;

inline ll exgcd(ll a, ll b, ll &x, ll &y){
	if(!b) {x = 1, y = 0; return a;}
	ll d = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return d;
}

inline bool solve(ll a, ll b, ll c, ll &ans){
	ll x, y, d = exgcd(a, b, x, y);
	if(c % d) return 0;
	ll looplen = b / d;
	ans = x * (c / d) % looplen; // Enlarge
	if(ans < 0) ans += looplen;
	return 1;
}

int main(){
	while(scanf("%lld%lld%lld%d", &A, &B, &C, &k) != EOF && k){
		if(!solve(C, 1LL << k, B - A, ans)) puts("FOREVER");
		else printf("%lld\n", ans);
	}
	return 0;
}

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OPEN AT 2017.12.10

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