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 < 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
推論1：方程ax=b(mod n)對于未知量x有解，當(dāng)且僅當(dāng)gcd(a,n) | b。
推論2：方程ax=b(mod n)或者對模n有d個不同的解，其中d=gcd(a,n)，或者無解。
定理1：設(shè)d=gcd(a,n)，假定對整數(shù)x和y滿足d=ax+by(比如用擴(kuò)展Euclid算法求出的一組解)。如果d | b，則方程ax=b(mod n)有一個解x0滿足x0=x*(b/d) mod n 。特別的設(shè)e=x0+n，方程ax=b(mod n)的最小整數(shù)解x1=e mod (n/d)，最大整數(shù)解x2=x1+(d-1)*(n/d)。
定理2：假設(shè)方程ax=b(mod n)有解，且x0是方程的任意一個解，則該方程對模n恰有d個不同的解(d=gcd(a,n))，分別為：xi=x0+i*(n/d) mod n 。
以上定理的具體證明見《算法導(dǎo)論》。
#include <iostream>
using namespace std;
long long ext_gcd(long long a,long long b,long long &x,long long &y)
{
long long t,ret;
if(!b){
x=1,y=0;
return a;
}
ret=ext_gcd(b,a%b,x,y);
t=x,x=y,y=t-a/b*y;
return ret;
}
long long modular_linear(long long a,long long b,long long n){
long long d,e,x,y;
d=ext_gcd(a,n,x,y);
if(b%d)
return -1;
e=x*(b/d)%n+n;
return e%(n/d);
}
int main(){
long long d,a,b,c,k;
while(scanf("%lld %lld %lld %lld",&a,&b,&c,&k),a||b||c||k){
d=modular_linear(c,b-a,1LL<<k);
if(d==-1)
puts("FOREVER");
else
printf("%lld\n",d);
}
return 0;
}