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最新推荐文章于 2022-11-19 08:50:17 发布

UnICorN21

最新推荐文章于 2022-11-19 08:50:17 发布

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分类专栏： ACM_Steps 文章标签： C++ hdu

本文链接：https://blog.csdn.net/unicornet/article/details/8898071

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Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input

Sample Output

好吧，我不该轻视这题的，本来在各种地方看过n多次了，以为一次可以ac，结果各种特殊情况(其实按正常思路就该考虑到那些的--！)wa了2次TAT，唉……废话不多说，本题经典贪心，上代码。

#include <iostream>
#include <fstream>
#include <iomanip>
#include <algorithm>
using namespace std;
// greedy
struct Food{
	int j, f;
}food[1005];
bool cmp(Food a, Food b){return a.j * 1.0 / a.f > b.j * 1.0 / b.f;}
int main(){
	int m, n, p; 
	double ans;
	//ifstream fin("data.in");
	while (p = ans= 0, cin >> m >> n){
		if (m == -1 && n == -1) break;
		if (!n) {cout << "0.000" << endl; continue;}
		for (int i = 0; i < n; ++i) cin >> food[i].j >> food[i].f;
		sort(food, food + n, cmp);
		for (int i = 0; i < n; ++i) if (!food[i].f) {ans += food[i].j; food[i].j = 0;}
		while (m){
			if (m > food[p].f){m -= food[p].f; ans += food[p].j;}
			else{ans += food[p].j * 1.0 * m / food[p].f; m = 0;}
			++p;
		}
		cout << setprecision(3) << fixed << ans << endl;
	}
	//system("pause");
	return 0;
}