my扩展欧几里得

ll e_gcd(ll a,ll b,ll& x,ll& y){
if(b==0){
x=1;y=0;
return a;
}
ll ans = e_gcd(b,a%b,x,y);
ll tmp = x;
x=y;
y=tmp-a/b*y;
return ans;
}

ll  acl(ll a,ll b,ll c){
ll x,y;
ll gcd = e_gcd(a,b,x,y);
if(c%gcd!=0) return -1;
x *= c/gcd;
b /= gcd;
if(b<0) b=-b;
ll ans = x%b;
if(ans<0) ans+=b;
return ans;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define INT_MAX 2147483647
using namespace std;
typedef long long ll;
ll x,y;

ll abs(ll a){
if(a<0) return -a;
else return a;
}

ll e_gcd(ll a,ll b){
if(b==0){
x=1;y=0;
return a;
}
ll ans = e_gcd(b,a%b);
ll tmp = x;
x=y;
y=tmp-a/b*y;
return ans;
}

void  acl(ll a,ll b,ll c){
ll xx,yy,tx,ty;
int flag=0;
if(a<b){
swap(a,b);
flag = 1;
}
ll gcd = e_gcd(a,b);
x*=c/gcd; y*=c/gcd;//求出二元一次方程的解x,y
ll k =y*gcd/a,ans=INT_MAX;//利用y=a/gcd * t 求出 t 的值
for(ll i=k-5;i<k+5;i++){
//由于最后求得是 abs(x+b/gcd*t)+abs(y-a/gcd*) 的最小值，t是后者的最小，所以和最小在t的附近，遍历即可
xx = x + b/gcd*i;
yy = y - a/gcd*i;
if(abs(xx)+abs(yy)<ans){
ans=abs(xx)+abs(yy);
tx=xx;
ty=yy;
}
}
if(flag == 0){
cout << abs(tx) << ' ' << abs(ty);
}else{
cout << abs(ty) << ' ' << abs(tx);
}
cout << endl;

}

int  main(){
ll a,b,d;
while(scanf("%lld%lld%lld",&a,&b,&d)!=EOF){
if(a==0&&b==0&&d==0) break;
acl(a,b,d);
}

return 0;
}

 C Looooops

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define INT_MAX 2147483647
using namespace std;
typedef long long ll;
ll x,y;

ll e_gcd(ll a,ll b){
if(b==0){
x=1;y=0;
return a;
}
ll ans = e_gcd(b,a%b);
ll tmp = x;
x=y;
y=tmp-a/b*y;
return ans;
}

ll  acl(ll a,ll b,ll c){

ll gcd = e_gcd(a,b);
if(c%gcd!=0) return -1;
x*=(c/gcd);
b/=gcd;
if(b<0) b=-b;
ll ans = x%b;
if(ans<0) ans += b;
return ans;
}

int  main(){
ll a,b,c,k;
while(scanf("%lld%lld%lld%lld",&a,&b,&c,&k)!=EOF){
if(a==0&&b==0&&c==0&&k==0) break;
ll b1 = (ll)1<<(k);//注意使用强制转换
ll ans = acl(c,b1,b-a);
if(ans==-1)
cout << "FOREVER" ;
else
cout << ans;
cout << endl;
}

return 0;
}