description
现有两组数字，每组k个，第一组中的数字分别为：a1，a2，...，ak表示，第二组中的数字分别用b1，b2，...，bk表示。其中第二组中的数字是两两互素的。求最小的非负整数n，满足对于任意的i，n - ai能被bi整除。
analysis
- 根据题意可以知道
[ left{
egin{aligned}
n-a_1≡0(mod b_1)\
n-a_2≡0(mod b_2)\
......\
n-a_k≡0(mod b_k)
end{aligned}
ight.
]
-
对于(a-b≡0(mod c))，如果(c|a-b)，那么(amod c=b)
-
所以
[ left{
egin{aligned}
n≡a_1(mod b_1)\
n≡a_2(mod b_2)\
......\
n≡a_k(mod b_k)
end{aligned}
ight.
]
- 随便用哪种(CRT)都可以，但注意普通(CRT)的乘法每处都要模
code
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 100005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
using namespace std;
ll a[MAXN],m[MAXN];
ll n;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline ll exgcd(ll a,ll b,ll &x,ll &y)
{
if (b==0){x=1,y=0;return a;}
ll ans=exgcd(b,a%b,y,x);
y-=x*(a/b);return ans;
}
inline ll mul(ll x,ll y,ll mod)
{
return (x*y-(ll)((long double)x/mod*y)*mod+mod)%mod;
}
inline ll CRT(ll a[],ll m[],ll n)
{
ll M=1,x,y,ans=0;
fo(i,1,n)M*=m[i];
fo(i,1,n)
{
ll tmp=M/m[i];exgcd(tmp,m[i],x,y);
ans=(ans+mul(mul(a[i],x,M),tmp,M))%M;
}
return (ans%M+M)%M;
}
inline ll EXCRT(ll a[],ll m[],ll n)
{
ll M=m[1],x,y,ans=a[1];
fo(i,2,n)
{
ll gcd=exgcd(M,m[i],x,y),tmp=(a[i]-ans%m[i]+m[i])%m[i];
if (tmp%gcd!=0)return -1;
x=mul(x,tmp/gcd,m[i]/gcd);
ans+=x*M,M*=m[i]/gcd,ans=(ans%M+M)%M;
}
return (ans%M+M)%M;
}
int main()
{
n=read();
fo(i,1,n)a[i]=read();
fo(i,1,n)m[i]=read();
printf("%lld
",CRT(a,m,n));
return 0;
}