poj 2115

题目链接 http://poj.org/problem?id=2115

题目意思就解这样一个同余方程 A+CX≡B（mod）M 这里 M = 1<<k 跟上一篇的做法差不多。用扩展欧几里得做

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <vector>

using namespace std;

const int maxn = 1000005;

typedef long long LL;

LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
   if(b == 0){
        x = 1;
        y = 0;
        return a;
   }
   LL r = ex_gcd(b,a%b,x,y);
   LL t = x;
      x = y;
      y = t - a/b*y;
      return r;
}
int main()
{
    LL i,ans,a,b,c,d,k,x,y;
    while(cin>>a>>b>>c>>k&&a+b+c+k){
      LL tmp = c;
           c = b - a;
           a = tmp;
           b = (LL)1 << k;
           LL d = ex_gcd(a,b,x,y);
           if(c%d!=0){
            printf("FOREVER
");
           }
           else{
            ans = x*c/d;
            LL t = b/d;
            ans = ((ans%t)+t)%t;
            printf("%lld
",ans);
           }

    }
    return 0;
}