中国剩余定理

 对于
$$(s):\{\begin{array}{llc}x\equiv a_1& (mod& m_1)\\ x\equiv a_2& (mod& m_2)\\ x\equiv a_3& (mod& m_3)\\ ...& & \\ x\equiv a_n& (mod& m_n)\end{array}$$
我们求一组通解
易知通解为
${\sum }_{i=1}^n{a}_i\cdot M_i\cdot t_i$
其中
$$M=\prod _{i=1}^n{m}_i$$

$$M_i=\frac{M}{m_i}$$
$$t_i\cdot M_i\equiv 1(modm)$$

容易看出来这样构造出的解周期为$M$，只要在最后的解中$\%M$即可
Code

#include<cstdio>
#include<iostream>
using namespace std;
struct poi
{
	
	int mod,res,m,nm;
	
	
}a[10005];
int n;
void exgcd(int a,int b,int &x,int &y)
{
	if(b==0)
	{
		x=1;
		y=0;
		return ;
	}
	exgcd(b,a%b,y,x);
	y-=a/b*x;
	return ;
}
int main()
{
	
	cin>>n;
	int kkk=1;
	for(int i=1;i<=n;i++)
	{
	cin>>a[i].mod>>a[i].res;	
	kkk*=a[i].mod;
	}
	for(int i=1;i<=n;i++)
	{
		a[i].m=kkk/a[i].mod;
		//cout<<a[i].m<<endl;
	}
	int total=0;
	for(int i=1;i<=n;i++)
	{
		int x,y;
		exgcd(a[i].m,a[i].mod,x,y);
		//cout<<x<<endl;
		a[i].nm=(x%a[i].mod+a[i].mod)%a[i].mod;
		//cout<<a[i].nm<<endl;
		total+=a[i].m*a[i].nm*a[i].res;
	}	
	total%=kkk;
	cout<<total;
}