POJ - 2976 Dropping tests && 0/1 分数规划

POJ - 2976 Dropping tests

你有 \(n\) 次考试成绩， 定义考试平均成绩为 $$\frac{\sum_{i = 1}^{n} a_{i}}{\sum_{i = 1}^{n} b_{i}}$$

你可以考虑放弃 \(K\) 次成绩， 求最大平均成绩 * 100

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小插曲： 被精度卡成喜羊羊

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0/1分数规划\(from\)人生导师

Solution

01分数规划（不是很）裸题， 在每次 \(check\) 时， 选取较大的 \(num - K + 1\) 次即可

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
using namespace std;
double RD(){
    double out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const int maxn = 2019;
int num, K;
double a[maxn], b[maxn],c[maxn];
bool check(double k){
	for(int i = 1;i <= num;i++){
		c[i] = a[i] - b[i] * k;
		}
	sort(c + 1, c + 1 + num);
	double temp = 0;
	for(int i = num;i > K;i--)temp += c[i];
	if(temp > 0)return 1;
	return 0;
	}
double search(double l, double r){
	double ans;
	while(r - l > 0.000000001){
		double mid = (l + r) / 2;
		if(check(mid))ans = mid, l = mid;
		else r = mid;
		}
	return ans;
	}
int main(){
	while(1){
		num = RD(), K = RD();
		if(!num && !K)return 0;
		for(int i = 1;i <= num;i++)a[i] = RD();
		for(int i = 1;i <= num;i++)b[i] = RD();
		printf("%.0f\n",100 * search(0, 1000000019));
		}
	}