POJ2115——C Looooops(擴充歐幾裡德+求解模線性方程)

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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

題目大意：

　　　　對於C的for(i=A ; i!=B ;i +=C)迴圈語句，問在k位儲存系統中迴圈幾次才會結束。

　　　　若在有限次內結束，則輸出迴圈次數。

　　　　否則輸出死迴圈。

解題思路：

　　　　有題意可知：設進行X次迴圈 A%2^K+(C*X)%2^K=B%2^K

　　　　　　-->(C*X)%(2^K)=(B-A)%(2^K)

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

演算法：

　　　　演算法導論上給出了 求解a=b (mod n)的虛擬碼 其中a,n為正整數，b為任意整數。

　　　　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)為所有可以的解，輸出其最小正整數解即可。

Code：

 1 /************************************************************************* 2     > File Name: poj2115.cpp 3     > Author: Enumz 4     > Mail: [email protected] 5     > Created Time: 2014年10月29日 星期三 13時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 10000023 #define LL long long24 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         else66             printf("FOREVER\n");67     }68     return 0;69 }

 

　　　　

POJ2115——C Looooops(擴充歐幾裡德+求解模線性方程)