poj 2115 C Looooops(扩展欧几里得)

来源：互联网发布：java反射获取私有属性编辑：程序博客网时间：2024/05/18 02:09

也是比较基础的吧，根据题意可以列出式子：Cx+2^k * y=B-A，x为要循环的次数，用扩展欧几里得求解即可。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<set>#include<stack>#include<cmath>#include<vector>#define inf 0x3f3f3f3f#define Inf 0x3FFFFFFFFFFFFFFFLL#define eps 1e-9#define pi acos(-1.0)using namespace std;typedef long long ll;ll ext_gcd(ll a,ll b,ll &x,ll &y){    if(b==0)    {        x=1;y=0;        return a;    }    ll d=ext_gcd(b,a%b,y,x);    y-=a/b*x;    return d;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    ll A,B,C,k,x,y,a,b,c,d;    while(cin>>A>>B>>C>>k)    {        if(A==0&&B==0&&C==0&&k==0) break;        k=(1LL)<<k;        a=C;b=k;c=B-A;        d=ext_gcd(a,b,x,y);        if(c%d!=0)            cout<<"FOREVER"<<endl;        else        {            c/=d;x*=c;b/=d;            x=(x%b+b)%b;            cout<<x<<endl;        }    }    return 0;}

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