poj 2115 C Looooops 【扩展欧几里得】

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20702		Accepted: 5596

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

题意：求解最小的y使得 A + C*y = B (mod 1<<k)。

思路：化简等式 (1<<k)*x + y*C = B-A。当且仅当(B-A) % gcd(1<<k, C) == 0才有解。

令temp = (B-A)/gcd(1<<k, C);

化简——(1<<k)*x/temp + y*C/temp = gcd(1<<k, C);

求出x和y，y *= temp。

由扩展欧几里得 定理三——若gcd(a, b) = d，则方程ax ≡ c (mod b)在[0, b/d - 1]上有唯一解。

求解最小的y时，只需对(1<<k) / gcd(1<<k, C)取余即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
LL A, B, C, k;
void exgcd(LL a, LL b, LL &d, LL &x, LL &y)
{
    if(b == 0) {d = a, x = 1, y = 0;}
    else
    {
        exgcd(b, a%b, d, y, x);
        y -= x * (a / b);
    }
}
void solve()
{
    B -= A;
    LL d, x, y;
    LL a = (1LL<<k);
    exgcd(a, C, d, x, y);
    if(B % d)
        printf("FOREVER\n");
    else
    {
        x *= (B / d); y *= (B / d);
        y = (y % (a / d) + a / d) % (a / d);
        printf("%lld\n", y);
    }
}
int main()
{
    while(scanf("%lld%lld%lld%lld", &A, &B, &C, &k) != EOF)
    {
        if(A == 0 && B == 0 && C == 0 && k == 0)
            break;
        solve();
    }
    return 0;
}