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
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 "); 57 else{ 58 int ans = (n-r)/t + 1; 59 if(r == 0) ans--; 60 printf("%d " , ans); 61 } 62 } 63 return 0; 64 }