链接：http://acm.hdu.edu.cn/showproblem.php?pid=1573

题意：求在小于等于N的正整数中有多少个X满足：X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod a[i] = b[i], … (0 < a[i] <= 10)。

思路：中国剩余定理的模板题，如果找不到这样的数或者最小的X大于N，输出零。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <ctype.h>
#include <algorithm>
#include <string>
#include <set>
#define PI acos(-1.0)
#define maxn 10005
#define INF 0x7fffffff
#define eps 1e-8
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
LL MOD;
LL extend_gcd(LL a, LL b, LL &x, LL &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    LL r=extend_gcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
LL inv(LL a,LL m)
{
    LL d,x,y;
    d=extend_gcd(a,m,x,y);
    if (d==1)
    {
        x=(x%m+m)%m;
        return x;
    }
    else return -1;
}
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}
bool merge(LL a1,LL m1,LL a2,LL m2,LL &a3,LL &m3)
{
    LL d=gcd(m1,m2);
    LL c=a2-a1;
    if(c%d)
        return false;
    c=(c%m2+m2)%m2;
    c/=d;
    m1/=d;
    m2/=d;
    c*=inv(m1,m2);
    c%=m2;
    c*=m1*d;
    c+=a1;
    m3=m1*m2*d;
    a3=(c%m3+m3)%m3;
    return true;
}
LL CRT_next(LL a[],LL m[],int n)
{
    LL a1=a[0],m1=m[0],a2,m2;
    for(int i=1;i<n;i++)
    {
        LL aa,mm;
        a2=a[i],m2=m[i];
        if(!merge(a1,m1,a2,m2,aa,mm))
            return -1;
        a1=aa;
        m1=mm;
    }
    MOD=m1;
    LL aa=(a1%m1+m1)%m1;
    if(aa==0)
    aa+=m1;
    return aa;
}
int main()
{
    int T;
    LL a[55],b[55];
    scanf("%d",&T);
    for(int ii=1; ii<=T; ii++)
    {
        int tot;
        LL t1;
        scanf("%I64d%d",&t1,&tot);
        for(int i=0; i<tot; i++)
            scanf("%I64d",&a[i]);
        for(int i=0; i<tot; i++)
            scanf("%I64d",&b[i]);
        if(tot==1)
        {
            if(b[0]==0)
                b[0]+=a[0];
            if(t1<b[0])
                printf("0\n");
            else
                printf("%I64d\n",(t1-b[0])/a[0]+1);
        }
        else
        {
            LL ans=CRT_next(b,a,tot);
            if(ans==-1)
                printf("0\n");
            else if(ans>t1)
                printf("0\n");
            else
                printf("%I64d\n",(t1-ans)/MOD+1);
        }
    }
    return 0;
}