C++_FatMouse' Trade(贪心)
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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
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
Sample Output
13.333
31.500
思路
获得每一个的价格,然后每次都拿最小的。一开始建立了三个数组,然后打算自己写一个排序算法,根据价格的值同时修改三个数组,后来受人指点,把三个值写成一个结构体就行了。
#include <iostream>#include <cstdio>#include <vector>#include <algorithm>#define N 1005//#define DEBUGusing namespace std;int m, n;double ans;struct Fat{ double price; int j, f;};vector<Fat> good(N);bool myCompare(Fat a, Fat b) { return a.price < b.price; }int main(){ while(~scanf("%d%d", &m, &n)) { ans = 0; if(m<0||n<0) break; for(int i = 0;i<n;i++) { scanf("%d%d", &good[i].j, &good[i].f); good[i].price = (double)good[i].f/(double)good[i].j; } sort(good.begin(), good.begin() + n, myCompare);#ifdef DEBUG for(int i= 0;i<n;i++) { cout << " J : " << good[i].j << "F : " << good[i].f << endl << "Price : " << good[i].price << endl; }#endif // DEBUG for(int i = 0;i<n;i++) { if(m<=0) break; if(m>=good[i].f) { ans+=good[i].j; m-=good[i].f; } else { ans+=(double)m/good[i].price; m = 0; } } printf("%.3lf\n", ans); } return 0;}
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