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自然万物都趋向从有序变得无序

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C Looooops POJ - 2115

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题目链接:点我

题目

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:

问题目中所给出的伪代码的循环要执行几次.

思路:

扩展欧几里得算法,即是在求满足c * x  + y *(1 << k) = b - a;的最小整数解,由定理可得,当不满足(b - a ) % gcd ( c, 1 << k) == 0时,方程一定不会有解,那么意味着将死循环,满足是我们将方程两边同时除了 gcd(c, 1 << k),解出x,再将x乘(b - a) / gcd(c, 1 << k) 即可.

代码:

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#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;

typedef long long LL;

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

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

}
void solve(LL a,LL b,LL c){
    LL g= gcd(a,b);
    if(c%g!=0){
        printf("FOREVER\n");
        return ;
    }
    LL x,y;
    exgcd(a,b,x,y);
    x=x*(c/g) %b;
   LL ans=(x%(b/g)+b/g) %(b/g);
    printf("%I64d\n",ans);
}

int main(){
    LL a,b,c,k;
    while(scanf("%I64d %I64d %I64d %I64d", &a,&b, &c,&k) &&(a+b+c+k)){
        solve(c,(LL)1<<k,b-a);
    }
    return 0;
}