poj 2115 C Looooops

C Looooops

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17304 Accepted: 4475

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

大致题意：
对于C的for(i=A ; i!=B ;i +=C)循环语句，问在k位存储系统中循环几次才会结束。若在有限次内结束，则输出循环次数。否则输出死循环。
解题思路：
题意不难理解，只是利用了 k位存储系统 的数据特性进行循环。例如int型是16位的，那么int能保存2^16个数据，即最大数为65535（本题默认为无符号），当循环使得i超过65535时，则i会返回0重新开始计数如i=65534，当i+=3时，i=1其实就是 i=(65534+3)%(2^16)=1有了这些思想，设对于某组数据要循环x次结束，那么本题就很容易得到方程：x=[(B-A+2^k)%2^k] /C 即 Cx=(B-A)(mod 2^k)  此方程为 模线性方程，本题就是求X的值。

代码如下：

#include<iostream>#include<stdio.h>using namespace std;typedef long long LL;LL a,b,c,k;int K;LL x,y,gcd;LL extend_gcd(LL a,LL b){    if(b==0)    {        y=0;x=1;gcd=a;return a;    }    extend_gcd(b,a%b);    LL temp=x;    x=y;    y=temp-(a/b)*y;}int main(){    LL i;    while(scanf("%lld%lld%lld%d",&a,&b,&c,&K)&&a+b+c+K)    {        LL r=b-a;        LL k=(LL)1<<K;        extend_gcd(c,k);        if(r%gcd!=0)        {printf("FOREVER\n");continue;}        LL temp=k/gcd;        x=((r/gcd*x)%temp+temp)%temp;        printf("%lld\n",x);    }
    return 0;}