非互质的中国剩余定理。。。用那个叫合并上升法的东西。。。orzorzorz。。。
#include <iostream> #include <sstream> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <climits> #define maxn 800005 #define eps 1e-6 #define mod 10007 #define INF 99999999 #define lowbit(x) (x&(-x)) //#define lson o<<1, L, mid //#define rson o<<1 | 1, mid+1, R typedef long long LL; using namespace std; void extend_gcd(int a, int b, int &d, int &x, int &y) { if(b == 0) { d = a, x = 1, y = 0; } else { extend_gcd(b, a%b, d, y, x), y -= x*(a/b); } } int a[maxn], b[maxn]; int main(void) { int _, a1, a2, b1, b2, tmp, i, n, m, ok, g, x, y; while(scanf("%d", &_)!=EOF) { while(_--) { scanf("%d%d", &m, &n); for(i = 0; i < n; i++) scanf("%d", &a[i]); for(i = 0; i < n; i++) scanf("%d", &b[i]); a1 = a[0], b1 = b[0], ok = 0; for(i = 1; i < n; i++) { a2 = a[i], b2 = b[i]; extend_gcd(a1, a2, g, x, y); if((b2-b1)%g) { //无解的情况 ok = 1; break; } tmp = a2/g; x = x*(b2-b1)/g; //x 的所有解 x = (x%tmp+tmp)%tmp; //找到最小正解x b1 = a1*x+b1; //更新b1 a1 = (a1*a2)/g; //更新a1 b1 = (b1%a1+a1)%a1; //找到最小正解b1 } //解出的答案是b1,循环是a1. if(ok || m<b1) printf("0\n"); else printf("%d\n", (m-b1)/a1+1-(b1==0 ? 1 : 0)); } } return 0; }