01分数规划

输入第一行一个整数T(1<=T<=10)
接下来有T组测试数据，对于每组测试数据，第一行输入两个数n和k(1<=k<=n<=100000)
接下来有n行，每行两个是a和b，代表这个物品的重量和价值

对于每组测试数据，输出对应答案，结果保留两位小数

1
3 2
2 2
5 3
2 1

0.75

对于样例来说，我们选择第一个物品和第三个物品，达到最优目的

01分数规划的模板题，重载小于号，直接二分ans即可。二分小技巧，double的时候防止精度不动，直接二分100次。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
const double eps = 1e-6;
struct Node {
    double a, b, c;
    bool operator<(const Node& b) const {
        return c > b.c;
    }
}a[N];
int n, m;

bool check(double x) {
    for (int i = 1; i <= n; ++i)    a[i].c = a[i].b - x * a[i].a;
    sort(a + 1, a + 1 + n);
    double ans = 0;
    for (int i = 1; i <= m; ++i)
        ans += a[i].c;
    return ans > eps;
}

int main() {
    int T = read();
    while (T--) {
        n = read(), m = read();
        for (int i = 1; i <= n; ++i)    a[i].a = read(), a[i].b = read();
        double l = eps, r = 1e9, mid;
        for (int i = 1; i <= 100; ++i) {
            mid = (r + l) / 2.0;
            if (check(mid))    l = mid;
            else r = mid;
        }
        printf("%.2f\n", mid);
    }
    return 0;
}