【数论】 HDOJ 1573 X问题

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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xb1 = a1*x+b1; //更新b1a1 = (a1*a2)/g; //更新a1b1 = (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;}

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