[POJ2976] Dropping tests

传送门：>Here<

题意：给出长度相等的数组a和b，定义他们的和为$\dfrac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}$。现在可以舍弃k对元素（一对即$a[i]和b[i]$），问最大的和是多少？

解题思路

01分数规划入门题（并没有学过，看到hy大佬在刷因此也去学了下）

问题可以转化为数组中的每个元素选或不选，也就可以认为每一个元素都乘上一个$x[i], \ x[i] ∈ \{0, 1\}$

因此问题可以转化为$ans = \dfrac{\sum\limits_{i = 1}^{n}a[i] * x[i]}{\sum\limits_{i = 1}^{n}b[i] * x[i]}$

将除法转化为加法$\sum\limits_{i = 1}^{n}a[i] * x[i] - ans * \sum\limits_{i = 1}^{n}b[i] * x[i] = 0$

合并得$\sum\limits_{i = 1}^{n}(a[i]-ans*b[i])*x[i] = 0$

当$x$数组的取值确定时，可以发现函数$f(r) = \sum\limits_{i = 1}^{n}(a[i]-r*b[i])*x[i]$是减函数，因此可以二分$r$。当前取到的$r$能够满足$\sum\limits_{i = 1}^{n}(a[i]-r*b[i])*x[i] \geq 0$即为可行，为了满足此条件，肯定要让选择的那些元素的和越大越好，因此可以建立数组$d[i] = a[i]-r*b[i]$并排序，选择最大的加起来验证是否大于等于0.

Code

long long

/*By DennyQi 2018.8.12*/
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define  r  read()
#define  Max(a,b)  (((a)>(b)) ? (a) : (b))
#define  Min(a,b)  (((a)<(b)) ? (a) : (b))
using namespace std;
typedef long long ll;
#define int long long
const int MAXN = 10010;
const int MAXM = 27010;
const int INF = 1061109567;
inline int read(){
    int x = 0; int w = 1; register int c = getchar();
    while(c ^ '-' && (c < '0' || c > '9')) c = getchar();
    if(c == '-') w = -1, c = getchar();
    while(c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar(); return x * w;
}
struct Score{
    int idx; double sc;
}s[MAXN];
int N,K;
int a[MAXN],b[MAXN];
double L,R,Mid,d[MAXN];
inline bool comp(const Score& a, const Score& b){
    return a.sc < b.sc;
}
inline bool judge(double _r){
    for(int i = 1; i <= N; ++i){
        d[i] = (double)((double)(a[i]) - (double)(1.0*_r*b[i]));
    }
    sort(d+1,d+N+1);
    double res = 0.0;
    for(int i = N; i > K; --i){
        res += d[i];
    }
    return res >= 0.0;
}
#undef int
int main(){
#define int long long
    for(;;){
        N = r, K = r;
        if(!N && !K) break;
        for(int i = 1; i <= N; ++i) a[i] = r;
        for(int i = 1; i <= N; ++i) b[i] = r;
        L = 0.000, R = 9999999999.999;
        while(R - L >= 1e-8){
            Mid = (L + R) / 2.000;
            if(judge(Mid)){
                L = Mid;
            }
            else{
                R = Mid;
            }
        }
        for(int i = 1; i <= N; ++i){
            s[i] = (Score){i, (double)((double)(a[i]) - (double)(1.0*L*b[i]))};
        }
        sort(s+1,s+N+1,comp);
        int fz=0,fm=0;
        for(int i = N; i > K; --i){
            fz += a[s[i].idx];
            fm += b[s[i].idx];
        }
        double rs = (double)fz/(double)fm;
        printf("%.0f\n", rs * 100);
    }
    return 0;
}