C Looooops
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14925 | Accepted: 3767 |
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.
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.
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
02
32766
FOREVER
Source
扩展gcd：
形如 ax≡n (mod b),可化成 ax + by = n, x有整数解的充分必要条件是 n % （a，b）==0，(a,b)表示gcd(a,b)
方程 a*x+b*y=n；我们可以先用扩展欧几里德算法求出一组x_0，y_0。
也就是 a * x_0 + b * y_0 =（a，b）；
然后两边同时除以（a，b），再乘以n。
这样就得到了方程 a * x_0 * n/（a，b）+ b * y_0 * n/（a，b）= n；
方程的一个解: x = x_0 * n/（a，b）, y = y_0 * n/（a，b）
其他的解都为 x_1,2...≡x+(b/(a,b))*i , y_1,2...≡y-(a/(a,b))*i (0 <=i<=(a,b)-1)。 i=0 时即为上面的一组解
总共有gcd(a,b)组解
令ff=b/(a,b),则x的最小非负整数解为 x = x_0 * n/（a，b）, X = ( x % ff + ff ) % ff ;
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long int LL;
LL egcd(LL a,LL b,LL& x,LL& y)
{
if(b==0)
{
x=1;y=0;
return a;
}
else
{
LL r=egcd(b,a%b,y,x);
y-=x*(a/b);
return r;
}
}
int main()
{
LL A,B,C,k;
while(cin>>A>>B>>C>>k)
{
if(A==0&&B==0&&C==0&&k==0) break;
LL x0,y0,q,pow;
pow=1LL<<k;
q=egcd(C,pow,x0,y0);
if((B-A)%q!=0) puts("FOREVER");
else
{
LL x=x0*(B-A)/q;
LL ff=pow/q;
LL X=(x%ff+ff)%ff;
cout<<X<<endl;
}
}
return 0;
}