- 题目描述：
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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.
- 输入：
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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.
- 输出：
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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.
- 样例输入：
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5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
- 样例输出：
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13.333
31.500
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题目的意思是：有m元钱，有n种物品，每种物品有J磅，总价值f元，可以使用0到f的任意价格购买相应磅的物品。要求输出用m元钱最多能买到多少磅的物品。
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输入第一行是两个数字m和n，分别代表有m元和n种物品，接下来n行分被代表每种物品的重量和价值。输入m和n为-1时程序结束。
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输出为买到物品的重量。
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思路：这个题就是典型的贪心算法，现在再看这个算法，其实的确十分厉害。后悔当初没有好好学习。
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思路就是每次都购买剩下物品种性价比最高的物品，直到物品被买完或者钱用光了。
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代码如下：
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#include <iostream> #include<algorithm> #include<string.h> #include<stdio.h> using namespace std; struct goods { double j; // 重量 double f; // 价值 double s; // 性价比 bool operator < (const goods &A) const { return s > A.s; } }buf[1000]; int main() { double m; int n; int i; while(scanf("%lf%d", &m, &n)!=EOF) { if(m==-1 && n==-1 ) break; for(i=0; i<n; i++) { scanf("%lf %lf", &buf[i].j, &buf[i].f); buf[i].s = buf[i].j / buf[i].f; } sort(buf, buf+n); int idx=0; //当前货物下标 double ans = 0; //总重量 while(m>0 && idx<n) //循环条件是既有物品剩余还有钱剩余 { if(m>buf[idx].f) { ans += buf[idx].j; m -= buf[idx].f; } else { ans += buf[idx].j * m / buf[idx].f; //要理解这一步的含义 m=0; } idx ++; } printf("%.3lf\n", ans); } return 0; }