POJ2115C Looooops

http://poj.org/problem?id=2115

k位储存特点，一旦溢出，那么就到第二个循环开始返回0重新计数。问题实际转化成a+cx=b（mod 2^k）跑多少圈能够重合。因为是k位无符号，所以直接就是2^k次方，0~2^k-1。刚好覆盖模的范围

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<math.h>
 4 #include<algorithm>
 5 using namespace std;
 6 typedef unsigned long long ll;
 7 ll ex_gcd(ll a, ll b, ll &x,ll &y)
 8 {
 9     if(!b){x=1,y=0;return a;}
10     ll d=ex_gcd(b,a%b,x,y);
11     ll tmp=x;
12     x=y,y=tmp-a/b*y;
13     return d; 
14 }
15 int main()
16 {
17     ll a,b,c,k,x,y;
18     while(~scanf("%llu%llu%llu%llu",&a,&b,&c,&k),(a+b+c+k))
19     {
20         ll tmp=c;//k位无符号
21         c=b-a;a=tmp;b=(ll)1<<k;
22         ll d=ex_gcd(a,b,x,y);
23         ll t=b/d;
24         if(c%d!=0) cout<<"FOREVER\n";
25         else 
26         {
27             x=(x*c/d%t+t)%t;
28             cout<<x<<'\n';
29         }                     
30     }    
31     return 0;
32 }

标签：tmp,return,gcd,llu%,ll,Looooops,include,POJ2115C
来源： https://www.cnblogs.com/Astronaut0142/p/15083908.html