2020 CCPC Wannafly Winter Camp Day1 F 乘法（二分）

题意：

给定长度为 $n$n 的序列 $A$A 和长度为 $m$m 的序列 $B$B，序列 $A$A 的第 $i \left(1\leq i\leq n\right)$i(1≤i≤n)项为 $A_i$Ai​序列 $B$B 的第 $i \left(1\leq i\leq m\right)$i(1≤i≤m)项为 $B_i$Bi​

构造 $n\times m$n×m 的矩阵 $C$C ，第 $i \left(1\leq i\leq n\right)$i(1≤i≤n) 行第 $i \left(1\leq i\leq m\right)$i(1≤i≤m) 列的元素 $C_{i,j}=A_i\cdot B_j$Ci,j​=Ai​⋅Bj​
给定整数 $K$K ，求矩阵 $C$C 中第 $K$K 大的元素。

问题等价于求最大的数 $ans$ans ，使得 $C$C 中大于或等于 $ans$ans 的元素数目大于或等于 $K$K .
对序列 $A,B$A,B 按符号划分后再排序，二分答案即可。

令 $c$c 和 $C_{i,j}$Ci,j​同阶。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll g = exgcd(b, a % b, x, y);
    ll t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

const int maxn = 1e5 + 5;

ll a[maxn], b[maxn];
ll n, m, k;

bool check(ll mid)
{
    ll num = 0;
    for (int i = 1; i <= m; i++)
    {
        if (b[i] == 0)
            num += mid < 0 ? n : 0;
        if (b[i] < 0)
            num += lower_bound(a + 1, a + n + 1, ceil((double)mid / b[i])) - (a + 1);
        if (b[i] > 0)
            num += n - ((upper_bound(a + 1, a + n + 1, floor((double)mid / b[i]))) - (a + 1));
    }
    return num <= k;
}

int main()
{
    slddd(n, m, k);
    k--;
    rep(i, 1, n)
        sld(a[i]);
    rep(i, 1, m)
        sld(b[i]);
    sort(a + 1, a + n + 1);
    sort(b + 1, b + m + 1);
    ll l = -1e13, r = 1e13;
    while (l + 1 < r)
    {
        ll mid = (l + r) >> 1;
        if (check(mid))
            r = mid;
        else
            l = mid;
    }
    pld(r);
    return 0;
}