POJ2115 C Looooops

 1 /*
 2  POJ2115 C Looooops
 3  http://poj.org/problem?id=2115
 4  扩展欧几里得
 5  题意:求x, s.t. (a+c*x)=b (mod 1<<k)
 6         <=> c*x=b-a (mod 1<<k)
 7  */
 8 #include <cstdio>
 9 #include <algorithm>
10 #include <cstring>
11 #include <cmath>
12 #include <vector>
13 #include <queue>
14 #include <iostream>
15 #include <map>
16 #include <set>
17 //#define test
18 using namespace std;
19 const int Nmax=1e6+7;
20 long long a,b,c,k,x,y;
21 long long ex_gcd(long long a,long long b,long long &x,long long &y)//solve x,y in a*x+b*y=ex_gcd(a,b,x,y)=gcd(a,b);
22 {
23     if(b==0)
24     {
25         x=1LL;
26         y=0LL;
27         return a;
28     }
29     long long ans=ex_gcd(b,a%b,x,y);
30     long long tmp=x;
31     x=y;
32     y=tmp-a/b*y;
33     return ans;
34     //x = x0 + (b/gcd)*t
35     //y = y0 – (a/gcd)*t
36      
37 }
38 long long get(long long a,long long m,long long c)//get x in a*x=c(mod m)
39 {
40     //we can solve x,y in a*x+b*y=c <=> c%gcd(a,b)==0
41     long long x,y;
42     long long gcd=ex_gcd(a,m,x,y);
43     if(c%gcd!=0)
44         return -1;//error
45     m/=gcd;
46     x*=c/gcd;
47     if(m<0)
48         m=-m;
49     long long ans=x%m;
50     while(ans<0)
51         ans+=m;
52     return ans;
53 }
54 
55 
56 int main()
57 {
58     #ifdef test
59     #endif
60     //freopen("poj2115.in","r",stdin);
61     while(1)
62     {
63         scanf("%lld%lld%lld%lld",&a,&b,&c,&k);
64         if(!a&&!b&&!c&&!k)
65             break;
66         long long d=b-a;
67         //if(d<0)//即使是负数也不用考虑,可以得出正确答案
68             //d+=(1LL<<k);
69         long long ans=get(c,(1LL<<k),d);
70         if(ans==-1)
71             printf("FOREVER\n");
72         else
73             printf("%lld\n",ans);
74     }
75     return 0;
76 }

 

posted @ 2017-04-13 08:37  BBBob  阅读(105)  评论(0编辑  收藏