Total Submission(s): 3295 Accepted Submission(s): 1068
接下来两行。每行各有M个正整数。分别为a和b中的元素。
求解x在n的范围内的数量。由于全部的ai不是互质的,所以不能直接用中国剩余定理,
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define LL __int64 LL t , n , m , d , x , y , i , bb , aa , flag ; void gcd(LL a,LL b) { if(b == 0) { d = a ; x = 1 ; y = 0 ; } else { gcd(b,a%b); swap(x,y); x = -x ; y = -y ; y += (a/b)*x ; } return ; } LL a[30] , b[30] ; int main() { scanf("%I64d", &t); while(t--) { scanf("%I64d %I64d", &n, &m); for(i = 0 ; i < m ; i++) scanf("%I64d", &a[i]); for(i = 0 ; i < m ; i++) scanf("%I64d", &b[i]); aa = a[0] ; bb = b[0] ; flag = 1 ; for(i = 1 ; i < m ; i++) { gcd(aa,a[i]); if( (b[i]-bb)%d != 0 ) flag = 0 ; if( flag ) { x = (b[i]-bb)/d*x ; y = a[i] / d ; x = ( x%y + y )%y ; bb = bb + x * aa ; aa = aa*a[i]/d ; } } gcd(1,aa); if( bb%d != 0 ) flag = 0 ; if( flag ) { x = ( bb/d )*x ; y = aa / d ; x = (x % y + y) % y ; } if( flag == 0 || x > n ) printf("0\n"); else { if( !x ) printf("%I64d\n", (n-x)/y ); else printf("%I64d\n", (n-x)/y+1 ); } } return 0; }