题意描述
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.
思路
一道贪心的题目，我们可以按他的收益率来排序，每次选收益率最高的那一个。通过看其他人的做法，学会了自定义排序的方法。
代码语言：javascript复制bool cmp(Node x,Node y){
return x.rate>y.rate;
}通过重写cmp函数，可以实现对node结构体按收益率降序排序。
AC代码
代码语言：javascript复制#include<vector>
#include<iostream>
#include<cstring>
#include<queue>
#include<set>
#include<algorithm>
#define IOS ios::sync_with_stdio(false); cin.tie(0);
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
const int N=1005;
const int INF=0x3f3f3f3f;
int m,n;
int f[N],j[N];
typedef struct Node{
int j;
int f;
double rate;
};
Node node[N];
bool cmp(Node x,Node y){
return x.rate>y.rate;
}
int main() {
IOS;
while(cin>>m>>n&&m!=-1&&n!=-1){
for(int i=0;i<n;i ){
int a,b;
cin>>a>>b;
node[i].j=a,node[i].f=b;
node[i].rate=(double)a/b;
}
sort(node,node n,cmp);
double sum=0;
for(int i=0;i<n;i ){
if(m>node[i].f){
sum =node[i].j;
m-=node[i].f;
}else{
sum =node[i].rate*m;
m=0;
}
}
printf("%.3fn",sum)
}
return 0;
}
