POJ2115(扩展欧几里得)

C Looooops

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23700		Accepted: 6550

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 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0

Sample Output

0
2
32766
FOREVER

由题意易得(a+cx)%2^k==b，求x最小值。可得同余方程c*x=(b-a)mod2^k。

 //2016.8.17
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #define ll long long 

 using namespace std;

 ll ex_gcd(ll a, ll b, ll& x, ll& y)//扩展欧几里得
 {
     if(b==)
     {
         x = ;
         y = ;
         return a;
     }
     ll ans = ex_gcd(b, a%b, x, y);
     ll tmp = x;
     x = y;
     y = tmp-(a/b)*y;
     return ans;
 }

 int main()
 {
     ll a, b, c, x, y, res, n;
     int k;
     while(scanf("%lld%lld%lld%d", &a, &b, &c, &k)!=EOF)
     {
         if(!a&&!b&&!c&&!k)
           break;
         n = (ll)<<k;
         res = ex_gcd(c, n, x, y);
         cout<<res<<endl<<x<<endl;
         if((b-a)%res!=)cout<<"FOREVER"<<endl;
         else
         {
             x = x*(b-a)/res%n;//方程ax=b-a(mod n)的最小解
             ll tmp = n/res;
             x = (x%tmp+tmp)%tmp;//最小正数解
             printf("%lld\n", x);
         }
     }

     return ;
 }

 #include <iostream>
 #define ll long long

 using namespace std;

 ll ex_gcd(ll a, ll b, ll& x, ll& y){
     if(b == ){
         x = ;
         y = ;
         return a;
     }
     ll ans = ex_gcd(b, a%b, x, y);
     ll tmpx = x;
     x = y;
     y = tmpx-a/b*y;
     return ans;
 }

 int main()
 {
     int a, b, c, k;
     while(cin>>a>>b>>c>>k){
         if(!a&&!b&&!c&&!k)break;
         ll x, y;
         ll A = c;
         ll B = b-a;
         ll n = 1LL<<k;
         ll gcd = ex_gcd(A, n, x, y);
         if(B%gcd != )
             cout<<"FOREVER"<<endl;
         else{
             x = (x*(B/gcd))%n;
             x = (x%(n/gcd)+n/gcd)%(n/gcd);
             cout<<x<<endl;
         }
     }
     return ;
 }