【HDU1573】【扩展CRT模板题】X问题

Description

求在小于等于$N$的正整数中有多少个$X$满足：
$X \bmod a_0 = b_0$
$X \bmod a_1 = b_1$
$X \bmod a_2 = b_2$
$\dots$
$X \bmod a_i = b_i$
$\dots$$(0 < a[i] \le 10)$

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Solution

扩展CRT模板题。
考虑将同余方程两两合并，对于：
$x\equiv c_1\pmod {m_1}$
$x\equiv c_2\pmod {m_2}$
我们将其转化成：
$x=c_1+m_1x_1$
$x=c_2+m_2x_2$
联立、整理得：
$m_1x_1-m_2x_2=c_2-c_1$
这个方程可以用扩欧来解。
将解出来的$x_1$带入$x=c_1+m_1x_1$，求得$x$，则合并得到的方程为：
$y\equiv x\pmod{lca(m_1,m_2)}$
注意：这里的$x$为带入$x_1$求出来的值，$y$为方程的未知数！！

坑：注意处理$0$，题目中要求的是正整数解，$0$不能算！！

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Code

/************************************************
 * Au: Hany01
 * Date: May 27th, 2018
 * Prob: [HDU1573] X问题
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register LL i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register LL i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register LL i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline LL read()
{
    register LL _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

LL n, m, k, A, B, C, Gcd, t1, t2, ns;

struct Equation
{
    LL c, m;
}Eq[13];

LL gcd(LL x, LL y, LL& a, LL& b)
{
    if (!x) { a = 0, b = 1; return y; }
    register LL ret = gcd(y % x, x, a, b),t = a;
    a = b - a * (y / x);
    b = t;
    return ret;
}

int main()
{
#ifdef hany01
    File("hdu1573");
#endif

    for (static LL T = read(); T --; )
    {
        n = read(), m = read(), ns = 0;
        For(i, 1, m) Eq[i].m = read();
        For(i, 1, m) Eq[i].c = read();
        Fordown(i, m, 2) {
            A = Eq[i - 1].m, B = -Eq[i].m, C = Eq[i].c - Eq[i - 1].c;

            if (C % (Gcd = __gcd(A, B))) { ns = 1; break; }
            k = C / Gcd, gcd(A, B, t1, t2), t1 *= k, t2 *= k;

            Eq[i - 1].c = t1 * Eq[i - 1].m + Eq[i - 1].c;
            Eq[i - 1].m = Eq[i - 1].m * Eq[i].m / __gcd(Eq[i - 1].m, Eq[i].m);
        }
        ((Eq[1].c %= Eq[1].m) += Eq[1].m) %= Eq[1].m;
        if (ns || Eq[1].c > n) puts("0");
        else printf("%lld\n", (n - Eq[1].c) / Eq[1].m + 1 + (Eq[1].c ? 0 : -1));
    }

    return 0;
}
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