《ACM程序设计》书中题目P(会贿赂猫的胖老鼠)
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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.
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
题目大意就是一只胖老鼠会用猫粮和猫换食物,每间房的猫“汇率”都不一样,问如何换才能得到最多的食物;
思路如下:
1.根据每间房需要的猫粮和能换的食物来进行计算,算出“汇率”;
2.对汇率进行排序,选出交换的优先级;
3.一个一个计算直到猫粮耗尽为止。
程序如下:
#include<bits/stdc++.h>using namespace std;struct Food { double j; double f; double a; };bool cmp(const Food &a, const Food &b){ return a.a>b.a;}int main(){ Food x[1000]; int m,n,i; double s; while(cin>>m>>n) { if(m==-1) break; s=0; for(i=0;i<n;i++) { cin>>x[i].j>>x[i].f; x[i].a=x[i].j/x[i].f; //计算每间房的汇率 } sort(x,x+i,cmp); //对汇率进行排序 for(i=0;i<n;i++) //计算结果 { if(m>=x[i].f) { m-=x[i].f; s+=x[i].j; } else { s+=(x[i].a*m); break; } } cout<<setiosflags(ios::fixed)<<setprecision(3)<<s<<endl; }}
要注意保留3位小数。
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