poj 2115 C Looooops

 

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29034		Accepted: 8350

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

 

题意：k位存储元能存储2^k个二进制信息，题目即要求Cx==B-A (mod 2^k)的最小解x，这是线性模方程,其一般形式为ax==b(mod m)，转化一下：

设gcd(a.m)=G

那么ax==b(mod m)就可以转化为

(a/G)x==b/G (mod m/G) 继续转化

x==(a/G)^(-1)*(b/G)(mod m/G)

这样只要求出(a/G)的逆元，x即可求解

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<set>
#include<string>
#include<queue>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
LL A, B, C, k;
LL a, b, n;
LL extgcd(LL a,LL b,LL &x,LL &y) {
    LL d = a;
    if (b != 0) {
        d = extgcd(b,a%b,y,x);
        y -= (a / b)*x;
    }
    else {
        x = 1; y = 0;
    }
    return d;
}

LL mod_inverse(LL a,LL m) {
    LL x, y;
    extgcd(a,m,x,y);
    return (m + x%m) % m;
}

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

LL mod_fun(LL a,LL b,LL m,LL g) {
    return (b / g)*mod_inverse(a / g, m / g) % (m / g);
}


int main() {
    while (scanf("%lld%lld%lld%lld",&A,&B,&C,&k)&&(A||B||C||k)) {
        a = C;
        b = B - A;
        n = 1LL << k;
        LL g = gcd(a, n);
        if (b%g != 0) { puts("FOREVER"); continue; }
        if (b < 0)b += n;
        LL x = mod_fun(a, b, n, g);
        printf("%lld
",x);
    }
    return 0;
}