### 1 Description

Given two arrays of $$n$$ integers, $$num$$ and $$rem$$, find the minimum $$x$$ that satisfies $$x\ mod\ num_i=rem_i (i=1,…,n)$$.

### 2 Solution

Calculate the product of $$num$$ as $$prod$$.

Let $$nums_i$$ divide prod as $$pp_i$$.

And $$inv_i$$ is the modular multiplicative inverse of $$pp_i$$ with respect to $$num_i$$ that can be solved by extended Euclid’s Algorithm.

Then $$x=\sum_1^n(rem_i*pp_i*inv_i)%prod$$.

#include<bits/stdc++.h>
using namespace std;

int inv(int a, int m)
{
int m0 = m, t, q;
int x0 = 0, x1 = 1;
if (m == 1)
return 0;
while (a > 1)
{
q = a / m;
t = m;
m = a % m;
a = t;
t = x0;
x0 = x1 - q * x0;
x1 = t;
}
if (x1 < 0)
x1 = x1 + m0;
return x1;
}

int CRT(vector<int> num, vector<int> rem)
{
int prod = 1;
for (int i = 0; i < num.size(); i++)
{
prod = prod * num[i];
}
int res = 0;
int pp;
for (int i = 0; i < num.size(); i++)
{
pp = prod / num[i];
res = res + rem[i] * inv(pp, num[i])*pp;
}
return res % prod;
}

int main()
{
int n;
scanf("%d", &n);
vector<int> num(n);
vector<int> rem(n);
for (int i = 0; i < n; i++)
{
scanf("%d", &num[i]);
}
for (int i = 0; i < n; i++)
{
scanf("%d", &rem[i]);
}
printf("%d\n",CRT(num,rem));
system("pause");
return 0;
}


Result:

3
3 4 5
2 3 1
11