poj 2115 C Looooops(扩展欧几里德)

来源：互联网发布：了解算法看哪本书编辑：程序博客网时间：2024/06/06 14:08

C Looooops

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23143 Accepted: 6357

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(mod 2^k)中循环几次才会结束。

使用扩展欧几里德的方法，看不懂的请看我博客的相关知识

代码如下：

Memory: 412KTime: 0MSLanguage: G++Result: AcceptedSource Code#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;long long e_gcd(long long a, long long b, long long &x, long long &y){    if ( b == 0 )    {        x = 1;        y = 0;        return a;    }    long long ant = e_gcd(b, a%b, x, y);    long long temp = x;    x = y;    y = temp-a/b*y;    return ant;}int main(){    int a, b, c, d;    while ( ~scanf ( "%d %d %d %d", &a, &b, &c, &d )&&(a != 0||b != 0||c != 0||d != 0) )    {        long long x, y;                /*扩展欧几里德   相当于cx == ( b-a )( mod pow(2, d) )*/        long long ant = e_gcd(c, pow(2, d), x, y);                long long h = b-a;        if ( h%ant != 0 )            printf ( "FOREVER\n" );        else        {            x = x*h/ant;                        /*通解中正整数中最小的*/            long long s = pow(2,d)/ant;            x = (x%s+s)%s;                        printf ( "%lld\n", x );        }    }}

代码菜鸟，如有错误，请多包涵！！！

如有帮助记得支持我一下，谢谢！！！

0 0