HDU 3579 模余方程
/*HDU 3579不一定互质情况，神牛公式解析链接：http://yzmduncan.iteye.com/blog/1323599把推导过程翻译成代码即可... */#include<iostream>#include<cstdio>#include<cmath>using namespace std;typedef long long ll;
ll gcd(ll a,ll b){ if(b==0) return a; return gcd(b,a%b);}
ll kzgcd(ll a,ll b,ll&x,ll&y){ if(b==0) { x=1, y=0; return a; } ll d = kzgcd(b,a%b,x,y); ll k = y; y = x - a/b*y; x = k; return d;}
ll inv(ll a,ll b){ /// 求 a 的逆元 整数最小x ll x,y; ll k = kzgcd(a,b,x,y); if(k!=1) return -1; return (x%b+b)%b;}
bool merge(ll a1,ll n1,ll&a2,ll&n2){ ll d = gcd(n1,n2); ll c = a2 - a1; if(c%d) return false; ll k = inv(n1/d,n2/d)*c/d; ll a,n; n = n1/d*n2; a = (n1*k+a1)%(n1*n2/d); a2 = a; n2 = n; return true;}
ll china_reminder(ll k,ll*a,ll*b){ /// 余数，mod数 ll n1=b[1],a1=a[1]; for(ll i=2;i<=k;i++){ ll nn,aa; nn = b[i], aa = a[i]; if(!merge(a1,n1,aa,nn)) return -1; n1 = nn; a1 = aa; } return (a1%n1+n1)%n1;}
int main(){ ll t,a[1110],b[1100]; int f; cin>>f; for(int ca=1;ca<=f;ca++){ cin>>t; for(ll i=1;i<=t;i++) cin>>b[i]; for(ll i=1;i<=t;i++) cin>>a[i]; printf("Case %d: ",ca); ll sum = china_reminder(t,a,b); if(!sum){ /// 特判... sum=1; for(int i=1;i<=t;i++) sum = sum*b[i]/gcd(sum,b[i]); } printf("%lld\n",sum); }}
