题目：

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

题意：

Problem Description
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)。

Input
输入数据的第一行为一个正整数T，表示有T组测试数据。每组测试数据的第一行为两个正整数N，M (0 < N <= 1000,000,000 , 0 < M <= 10)，表示X小于等于N，数组a和b中各有M个元素。接下来两行，每行各有M个正整数，分别为a和b中的元素。

Output
对应每一组输入，在独立一行中输出一个正整数，表示满足条件的X的个数。

思路：

线性同余方程组模板题，贴两个模板
模板一：此模板要求同余方程组必须是如x≡ri(modmi)的形式，若给出的同余方程是aix≡ri(modmi)这样的形式，那么可以先求出逆元a−1imodmi，两边同时乘以逆元即可转换成如上形式

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 110, INF = 0x3f3f3f3f;

int M[N], R[N];

int gcd(int a, int b)
{
    return !b ? a : gcd(b, a%b);
}
int extgcd(int a, int b, int &x, int &y)
{
    int d = a;
    if(b)
    {
        d = extgcd(b, a%b, y, x);
        y -= (a/b) * x;
    }
    else x = 1, y = 0;
    return d;
}
int linear_congruence(int M[], int R[], int n, int num)
{
    int m = M[1], r = R[1];
    int x, y, flag = 1;
    for(int i = 2; i <= n; i++)
    {
        int d = gcd(m, M[i]), c = R[i] - r;
        if(c % d != 0)
        {
            flag = 0; break;
        }
        extgcd(m/d, M[i]/d, x, y);
        int tm = M[i] / d;
        x = ((c/d * x) % tm + tm) % tm;
        r = r + x*m;
        m = m/d * M[i];
        r %= m;
    }
    //if(r < 0) r += m;
    int ans = 0;
    if(num < r || flag == 0) ans = 0;
    else ans = (num - r)/m + 1;
    if(ans != 0 && r == 0) ans--;
    return ans;
}
int main()
{
    int t, n, m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= m; i++) scanf("%d", &M[i]);
        for(int i = 1; i <= m; i++) scanf("%d", &R[i]);
        printf("%d\n", linear_congruence(M, R, m, n));
    }
    return 0;
}

模板二：此模板可以解如aix≡ri(modmi)形式的同余方程组，模板一中的形式即ai=1的情况

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;
const int N = 110;
int A[N], B[N], M[N];

int gcd(int a, int b)
{
    return !b ? a : gcd(b, a % b);
}

int extgcd(int a, int b, int &x, int &y)
{
    int d = a;
    if(b)
    {
        d = extgcd(b, a%b, y, x);
        y -= (a / b) * x;
    }
    else x = 1, y = 0;
    return d;
}
int mod_inverse(int a, int m)
{
    int x, y;
    extgcd(a, m, x, y);
    return (m + x%m) % m;
}
P linear_congruence(const int *A, const int *B, const int *M, int len)
{
    int x = 0, m = 1;
    for(int i = 0; i < len; i++)
    {
        int a = A[i] * m, b = B[i] - A[i] * x, d = gcd(M[i], a);
        if(b % d != 0) return make_pair(0, -1);
        int t = b / d * mod_inverse(a / d, M[i] / d) % (M[i] / d);
        x = x + m * t;
        m *= M[i] / d;
    }
    return make_pair((x+m) % m, m);
}
int main()
{
    int t;
    int n, m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 0; i < m; i++) A[i] = 1;
        for(int i = 0; i < m; i++) scanf("%d", &M[i]);
        for(int i = 0; i < m; i++) scanf("%d", &B[i]);

        P p = linear_congruence(A, B, M, m);
        int ans = 0;
        if(p.second == -1) ans = 0;
        else if(n - p.first < 0) ans = 0;
        else ans = (n - p.first) / p.second + 1;
        if(p.first == 0 && ans) ans--;
        printf("%d\n", ans);
    }
    return 0;
}
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 1000 + 10;

ll extgcd(ll a, ll b, ll &x, ll &y)
{
    ll d = a;
    if(b != 0)
    {
        d = extgcd(b, a%b, y, x);
        y -= (a/b)*x;
    }
    else x = 1, y = 0;
    return d;
}
ll linear_congruence(ll m[], ll r[], ll n)
{
    ll M = m[1], R = r[1];
    for(ll i = 2; i <= n; i++)
    {
        ll x, y;
        int d = extgcd(M, m[i], x, y);
        ll c = r[i] - R;
        if(c % d) return -1;
        ll mod = m[i] / d;
        x = (c / d * x % mod + mod) % mod;
        R = R + x * M;
        M = M / d * m[i];
    }
    return R;
}
int main()
{
    int n;
    ll m[N], r[N];
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%lld%lld", &m[i], &r[i]);
    ll ans = linear_congruence(m, r, n);
    printf("%lld\n", ans);
    return 0;
}