洛谷P3868 [TJOI2009]猜数字

洛谷P3868 [TJOI2009]猜数字

TITLE

思路

中国剩余定理(CRT)
用龟速乘,注意a[i]=(a[i]%m+m)%m

CODE

#include<iostream> 
#include<cstdio>
using namespace std;
void exgcd(long long a,long long b,long long &x,long long &y)
{
    
    
	if(!b){
    
    x=1,y=0;return;}
	exgcd(b,a%b,x,y);
	long long tmp=x;x=y,y=tmp-a/b*y;
	return;
}
long long niyuan(long long s,long long m)
{
    
    
	long long x,y;
	exgcd(s,m,x,y);
	return (x%m+m)%m;
}
long long slowmul(long long x,long long y,long long m)
{
    
    
	long long ans=0;
	x=(x%m+m)%m,y=(y%m+m)%m;
	if(x<y)swap(x,y);
	for(;y;x=(x<<1)%m,y>>=1)
		if(y&1)ans+=x;
	return ans;
}
int main()
{
    
    
	long long n,lcm=1,x,y,i,ans=0,a[21],m[21];
	for(scanf("%lld",&n),i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(i=1;i<=n;i++)scanf("%lld",&m[i]),a[i]=(a[i]%m[i]+m[i])%m[i],lcm*=m[i];
	for(i=1;i<=n;i++)ans=(ans+slowmul(slowmul(lcm/m[i],a[i],lcm),niyuan(lcm/m[i],m[i]),lcm))%lcm;
	printf("%lld",(ans+lcm)%lcm);
	return 0;
}