HDU-1573-X问题（线性同余方程组）

链接：

https://vjudge.net/problem/HDU-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：

解线性同余方程组，得到(x+k*m leq n)。
解为(1+(n-x)/m)。
当x为0时答案要减一。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>

using namespace std;
typedef long long LL;
const int INF = 1e9;

const int MAXN = 1e6+10;

LL a, b, c, k, n, lcm;
int m;
LL M[20], A[20];

LL ExGcd(LL a, LL b, LL &x, LL &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    LL d = ExGcd(b, a%b, x, y);
    LL tmp = x;
    x = y;
    y = tmp - (a/b)*y;
    return d;
}

LL ExCRT()
{
    LL res = A[1], mod = M[1];
    LL x, y, gcd;
    for (int i = 2;i <= m;i++)
    {
        gcd = ExGcd(mod, M[i], x, y);
        if ((A[i]-res)%gcd)
            return -1;
        x = (x*(A[i]-res))/gcd;
        x = (x%(M[i]/gcd)+(M[i]/gcd))%(M[i]/gcd);
        res = res+mod*x;
        mod = (mod*M[i])/gcd;
        res %= mod;
    }
    lcm = mod;
    return res;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld%d", &n, &m);
        for (int i = 1;i <= m;i++)
            scanf("%lld", &M[i]);
        for (int i = 1;i <= m;i++)
            scanf("%lld", &A[i]);
        LL res = ExCRT();
        if (res == -1 || res > n)
            puts("0");
        else
        {
            printf("%lld
", (n-res)/lcm+(res!=0));
        }
    }

    return 0;
}