HDOJ 1573 X问题 (余数不互质的中国剩余定理)

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题意

求n以内有多少个数满足x%a[i]=b[i]。

思路

能发现我们可以用不互质的中国剩余定理求出来最小的x，然后(n-x)/lcm+1求个数
然后我就WA了好几发。。。注意求出的x为0且不是无解的时候最后的个数是(n-x)/lcm。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;typedef pair<LL, LL> pll;LL a[15], b[15], c[15];void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){    if (!b) {d = a, x = 1, y = 0;}    else    {        ex_gcd(b, a % b, y, x, d);        y -= x * (a / b);    }}LL inv(LL a, LL p){    LL x, y, d;    ex_gcd(a, p, x, y, d);    return d == 1 ? (x % p + p) % p : -1;}pll linear(LL A[], LL B[], LL M[], int n)      //求解A[i]x = B[i] (mod M[i]),总共n个线性方程组{    LL x = 0, m = 1;    for(int i = 0; i < n; i ++)    {        LL a = A[i] * m;        LL b = B[i] - A[i]*x;        LL d = __gcd(M[i], a);        if(b % d != 0)  return MP(0, -1);      //答案不存在，返回-1        LL t = b / d * inv(a / d, M[i] / d) % (M[i] / d);        x = x + m * t;        m *= M[i] / d;    }    x = (x % m + m) % m;    return MP(x, m);                           //返回的x就是答案，m是最后的lcm值}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int T;    scanf("%d", &T);    while (T--)    {        int n;        int m;        scanf("%d%d", &n, &m);        for (int i = 0; i < m; i++)            c[i] = 1;        for (int i = 0; i < m; i++)            scanf("%I64d", &a[i]);        for (int i = 0; i < m; i++)            scanf("%I64d", &b[i]);        pll ans = linear(c, b, a, m);        if (ans.first > n || ans.second == -1) printf("0\n");        else if (!ans.first) printf("%I64d\n", (n - ans.first) / ans.second);        else printf("%I64d\n", (n - ans.first) / ans.second + 1);    }    return 0;}

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