POJ2115——C Looooops(扩展欧几里德+求解模线性方程)

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C Looooops

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

题目大意：

　　　　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：

　　　　有题意可知：设进行X次循环 A%2^K+(C*X)%2^K=B%2^K

　　　　　　-->(C*X)%(2^K)=(B-A)%(2^K)

　　　　　　-->模线性方程 ax=b mod n的解。 其中a=C，b=B-A，n=2^K.

算法：

　　　　算法导论上给出了 求解a=b (mod n)的伪代码 其中a,n为正整数，b为任意整数。

　　　　MODULAR_LINEAR_EQUATION_SOLVER(a,b,n)

　　　　(d,x‘,y‘)=EXTENDED_EUCLID(a,n)

　　　　if (d|b)

　　　　　　x0=x‘(b/d) mod n

　　　　　　for i=0 to d-1

　　　　　　　　print (x0+i(n/d)) mod n

　　　　else

　　　　　　print "no solutions"

　　　　其中x0+i(n/d)为所有可以的解，输出其最小正整数解即可。

Code：

 1 /*************************************************************************
 2     > File Name: poj2115.cpp
 3     > Author: Enumz
 4     > Mail: 369372123@qq.com
 5     > Created Time: 2014年10月29日 星期三 13时13分46秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<cstdio>
10 #include<cstdlib>
11 #include<string>
12 #include<cstring>
13 #include<list>
14 #include<queue>
15 #include<stack>
16 #include<map>
17 #include<set>
18 #include<algorithm>
19 #include<cmath>
20 #include<bitset>
21 #include<climits>
22 #define MAXN 100000
23 #define LL long long
24 using namespace std;
25 LL extended_gcd(LL a,LL b,LL &x,LL &y)
26 {
27     LL ret,tmp;
28     if (b==0)
29     {
30         x=1,y=0;
31         return a;
32     }
33     ret=extended_gcd(b,a%b,x,y);
34     tmp=x;
35     x=y;
36     y=tmp-a/b*y;
37     return ret;
38 }
39 
40 int main()
41 {
42     LL A,B,C,k;
43     while (cin>>A>>B>>C>>k)
44     {
45         if (A==0&&B==0&&C==0&&k==0) break;
46         LL mod=1;
47         while (k--)
48             mod*=2;
49         LL x,y;
50         LL a=C,b=B-A;
51         LL d=extended_gcd(a,mod,x,y);
52         //cout<<a<<" "<<x<<" "<<mod<<" "<<y<<endl;
53         //cout<<d<<endl;
54         if (b/d*d==b)
55         {
56             LL x0=x*(b/d)%mod;
57             if(x0<0) x0+=mod;
58             //cout<<x0<<endl;
59             long long Min=x0;
60             for (int i=0;i<=d-1;i++)
61                 if ((x0+i*(mod/d))%mod>=0)
62                     Min=min(Min,(x0+i*(mod/d))%mod);
63             cout<<Min<<endl;
64         }
65         else
66             printf("FOREVER\n");
67     }
68     return 0;
69 }

POJ2115——C Looooops(扩展欧几里德+求解模线性方程)

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原文地址：http://www.cnblogs.com/Enumz/p/4060583.html