poj 2115 C Looooops 扩展欧几里德算法

发布时间:2016-12-7 5:47:38 编辑:www.fx114.net 分享查询网我要评论
本篇文章主要介绍了"poj 2115 C Looooops 扩展欧几里德算法",主要涉及到poj 2115 C Looooops 扩展欧几里德算法方面的内容,对于poj 2115 C Looooops 扩展欧几里德算法感兴趣的同学可以参考一下。

C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15305   Accepted: 3901 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 < 2k) modulo 2k.  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 < 2k) 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: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。 令a=C b=B-A n=2^k 那么原模线性方程变形为: ax=b (mod n) 该方程有解的充要条件为 gcd(a,n) | b ,即 b% gcd(a,n)==0 令d=gcd(a,n) 有该方程的 最小整数解为 x = e (mod n/d) 其中e = [x0 mod(n/d) + n/d] mod (n/d) ,x0为方程的最小解 那么原题就是要计算b% gcd(a,n)是否为0,若为0则计算最小整数解,否则输出FOREVER #include <iostream> #include <cstring> #include <cstdio> using namespace std; long long A,B,C,K; long long ex_gcd(long long a,long long &x,long long b,long long &y) { if(b==0) { x = 1; y = 0; return a; } long long d = ex_gcd(b,x,a%b,y); long long tmp = x; x = y; y = tmp - (a/b)*y; return d; } int main() { long long mod, X, Y; while(cin>>A>>B>>C>>K) { if(A==0&&B==0&&C==0&&K==0) break; mod = ((long long)1<<K); long long d = ex_gcd(C, X, mod, Y); long long T = B-A; if(T%d!=0) puts("FOREVER"); else { X = X*T/d; long long ans = (X%mod+mod)%mod; ans = ans%(mod/d); cout<<ans<<endl; } } return 0; } 算法参考:http://blog.csdn.net/lyy289065406/article/details/6648546