POJ 2115 C Looooops(单变元模线性方程)

C Looooops

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20180 Accepted: 5422

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 2 163 7 2 167 3 2 163 4 2 160 0 0 0

Sample Output

0232766FOREVER

求 Cx - (2^k)y = B-A中x的最小值

即求 ax - by = c;

先用扩展欧几里得求出  ax - by = gcd(a,b)的解(x,y)

如果c不能整除gcd(a,b)则无解

ax - by = gcd(a,b) 等式两边同时* c/gcd(a,b)得

ax * c/gcd(a,b) - by * c/gcd(a,b) = c

最后的解为 x * c/(gcd) 对 b/gcd(a,b)取余

#include <iostream>#include <cstdio>#include <cstring>#include <stdlib.h>#include <algorithm>#include <queue>#include <map>#include <cmath>#define N 1001#define ll long long int#define eps 1e-8#define PI acos(-1.0)#define inf 0x3f3f3f3fusing namespace std;ll x, y;ll gcd(ll a,ll b){    if(b == 0)    {        x = 1;        y = 0;        return a;    }    ll c = gcd(b, a%b);    ll t = x;    x = y;    y = t - (a/b)*y;    return c;}int main(){    ll A, B, C, k;    while(~scanf("%I64d%I64d%I64d%I64d", &A, &B, &C, &k))    {        if(!A && !B && !C && !k)            break;        ll a = C;        ll b = pow(2,k);        ll c = B - A;        ll gg = gcd(a, b);        if(c%gg)        {            printf("FOREVER\n");            continue;        }        x = x * (c/gg) ;      // ax - by = gg 变为  ax * c/gg - by * c/gg = c        x = (x%(b/gg) + b/gg) % (b/gg);   // x = x+k*( b/gg ) 求最小解        printf("%I64d\n",x);    }    return 0;}