HDU 1573 解同余模线性方程组

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 using namespace std;
 5 
 6 #define N 15
 7 int r[N] , m[N];
 8 
 9 int ex_gcd(int a , int &x , int b , int &y)
10 {
11     if(b == 0){
12         x = 1 , y = 0;
13         return a;
14     }
15     int ans = ex_gcd(b , x , a%b , y);
16     int t = x;
17     x = y , y = t - (a/b)*y;
18     return ans;
19 }
20 
21 int mod_line(int n , int &t)
22 {
23     int rr = r[0] , x , y;
24     t = m[0];
25     for(int i=1 ; i<n ; i++){
26         int del = r[i] - rr;
27         int g = ex_gcd(t , x , m[i] , y);
28         if(del % g != 0)
29             return -1;
30         int Mod = m[i] / g;
31         x = (((x*del/g)%Mod)+Mod)%Mod;
32 
33         rr = rr + t*x;
34         t = t*m[i] / g; //求二者最小公倍数，更新模项
35         rr %= t;
36     }
37     return rr;
38 }
39 
40 int main()
41 {
42    // freopen("a.in" , "r" , stdin);
43     int T;
44     scanf("%d" , &T);
45     while(T--)
46     {
47         int n , M;
48         scanf("%d%d" , &n , &M);
49         for(int i=0 ; i<M ; i++)
50             scanf("%d" , m+i);
51         for(int i=0 ; i<M ; i++)
52             scanf("%d" , r+i);
53 
54         int t;
55         int r = mod_line(M , t);
56         if(r>n || r == -1) printf("0\n");
57         else{
58             int ans = (n-r)/t + 1;
59             if(r == 0) ans--;
60             printf("%d\n" , ans);
61         }
62     }
63     return 0;
64 }