HDU 1573 解同余模线性方程组

题目意思很直接就是一道裸的解线性同余模方程组的题目

 #include <cstdio>
 #include <cstring>

 using namespace std;

 #define N 15
 int r[N] , m[N];

 int ex_gcd(int a , int &x , int b , int &y)
 {
     if(b == ){
         x =  , y = ;
         return a;
     }
     int ans = ex_gcd(b , x , a%b , y);
     int t = x;
     x = y , y = t - (a/b)*y;
     return ans;
 }

 int mod_line(int n , int &t)
 {
     int rr = r[] , x , y;
     t = m[];
     for(int i= ; i<n ; i++){
         int del = r[i] - rr;
         int g = ex_gcd(t , x , m[i] , y);
         if(del % g != )
             return -;
         int Mod = m[i] / g;
         x = (((x*del/g)%Mod)+Mod)%Mod;

         rr = rr + t*x;
         t = t*m[i] / g; //求二者最小公倍数，更新模项
         rr %= t;
     }
     return rr;
 }

 int main()
 {
    // freopen("a.in" , "r" , stdin);
     int T;
     scanf("%d" , &T);
     while(T--)
     {
         int n , M;
         scanf("%d%d" , &n , &M);
         for(int i= ; i<M ; i++)
             scanf("%d" , m+i);
         for(int i= ; i<M ; i++)
             scanf("%d" , r+i);

         int t;
         int r = mod_line(M , t);
         if(r>n || r == -) printf("0\n");
         else{
             int ans = (n-r)/t + ;
             if(r == ) ans--;
             printf("%d\n" , ans);
         }
     }
     return ;
 }