BZOJ 5418: [Noi2018]屠龙勇士

\[ \left\{\begin{array}{ccc} {xt_1} & {\equiv} & {a_{1}\pmod {p_{1}}} \\ {xt_2} & {\equiv} & {a_{2}\pmod {p_{2}}} \\ {} & {\vdots} & {} \\ {xt_n} & {\equiv} & {a_{k}\pmod {p_{n}}} \\ \end{array}\right. \]
\[ \left\{\begin{array}{ccc} {xt_1} & \geq & a_{1} \\ {xt_2} & \geq & a_{2} \\ {} & {\vdots} & \\ {xt_n} & \geq & a_{k} \\ \end{array}\right. \]
求出一个 \(x\) 满足上面两个方程组
首先用一次扩展欧几里得求出每个 \(x\) 的最小正整数解及其模数
再EXCRT合并，EXCRT的本质也是扩展欧几里得...
所以这是一道欧几里得算法练习题...
第二个不等式方程直接解就行

#include 
typedef long long ll;

namespace IO {
    const int MAXSIZE = 1 << 20;
    char buf[MAXSIZE], *p1, *p2;
    #define gc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin), p1 == p2) ? EOF : *p1++)
    void read() {}
    template
    inline void read(T &x, T2 &... oth) {
        x = 0; T f = 1; char ch = gc();
        while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = gc(); }
        while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = gc(); }
        x *= f;
        read(oth...);
    }
}

const int N = 1e5 + 7;
const ll INF = 0x3f3f3f3f3f3f3f3f;
std::multiset st;
ll a[N], r[N], p[N];
int n, m;

inline ll mul(ll a, ll b, ll mod) {
    a %= mod;
    if (a < 0) a += mod;
    b %= mod;
    if (b < 0) b += mod;
    return (a * b - (ll)((long double)a / mod * b) * mod + mod) % mod;
}

ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    ll g = exgcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - a / b * y;
    return g;
}

void equ(ll a, ll b, ll c, ll &x, ll &y) {
    ll g = exgcd(a, b, x, y);
    if (c % g) {
        x = y = -INF;
        return;
    }
    a /= g, b /= g, c /= g;
    x = (x % b + b) % b;
    x = mul(x, c, b);
    y = b;
}

bool merge(int i, int j) {
    ll x, y;
    equ(p[i], p[j], a[j] - a[i], x, y);
    if (x == -INF) return 0;
    a[j] = x * p[i] + a[i]; p[j] = y * p[i];
    return 1;
}

int main() {
    int T;
    IO::read(T);
    while (T--) {
        st.clear();
        IO::read(n, m);
        for (int i = 1; i <= n; i++)
            IO::read(a[i]);
        for (int i = 1; i <= n; i++)
            IO::read(p[i]);
        for (int i = 1; i <= n; i++)
            IO::read(r[i]);
        for (int i = 1; i <= m; i++) {
            ll x;
            IO::read(x);
            st.insert(x);
        }
        bool flag = 0;
        ll least = 0;
        for (int i = 1; i <= n; i++) {
            std::multiset::iterator it = st.upper_bound(a[i]);
            if (it != st.begin()) it--;
            ll v = *it, x, y; st.erase(it);
            st.insert(r[i]);
            equ(v, p[i], a[i], x, y);
            if (a[i] % v) least = std::max(least, a[i] / v + 1);
            else least = std::max(least, a[i] / v);
            //printf("%lld %lld %lld\n", v, x, y);
            if (x == -INF) {
                flag = 1;
                break;
            }
            a[i] = x; p[i] = y;

            if (i > 1) {
                if (!merge(i - 1, i)) {
                    flag = 1;
                    break;
                }
            }
        }
        if (flag) {
            puts("-1");
        } else {
            ll ans = a[n];
            if (a[n] < least) {
                least -= a[n];
                if (least % p[n])
                    least = (least / p[n] + 1) * p[n];
                ans = a[n] + least;
            }
            //printf("%lld %lld %lld\n", a[n], p[n], least);
            printf("%lld\n", ans);
        }
    }
}