poj 2115

模線性方程

需要注意的是mod 是間距也就是 ax+by =c ,  其中的 b/gcd(a,b)

模板題不需要多說

#include<iostream>
#include<cstdio>
using namespace std;
#define LL long long
inline  LL exgcd(LL a, LL b, LL &x, LL &y){/// ax + by = gcd(a, b);
    LL r;
    if(b){
        r =  exgcd(b, a%b, x, y);
        LL k = x;
        x = y;
        y = k - a/b*y;
    }
    else{
         y = (x = 1) - 1;
         return a;
    }
    return r;
}
LL mod_line(LL a, LL b, LL n)/// ak≡b(mod n) , ax + ny = b  x = x1*b/d * i(1...gcd(a, n));
/// k最小爲 (x1*b/d)%n/d
{
    LL x, y;
    LL d = exgcd(a, n, x, y);
    if(b%d){
        return -1;
    }
    else{
        x = (x*b/d) % n;
        LL mod = n/d;
        return (x%mod + mod) %mod;
    }
}
int main(){
    LL x, y, m, n, L;
    while(scanf("%lld%lld%lld%lld", &x, &y, &m, &n), x+y+m+n){
        /// x + k*m = y + k*n  k*(m - n) = y - x
        LL a = m;
        LL b = (y - x);
        L = (LL)(1LL<<n);
        LL k = mod_line(a, b, L);
        if(k == -1){
            printf("FOREVER\n");
        }
        else{
            printf("%lld\n", k);
        }
    }
    return 0;
}