FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 55094 Accepted Submission(s): 18478
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.
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
Author
CHEN, Yue
解题思路:
这是一个简单的贪心的题目，我感觉做贪心的题的时候需要注意的情况就是排序，只要抓住按照什么排序的话，
贪心就迎刃而解了，那这个题的话，就是按照J[i] / F[i]从大到小排序，只要排完序了，在根据题意进行一下计算
那不就是ok了，
上代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 100000+5;
const int mod = 1000000007;
const double eps = 1e-7;
struct Node
{
double J, F, ave;
}arr[maxn];
double cmp(Node a, Node b)
{
return a.ave > b.ave;
}
int main()
{
int m, n;
while(cin>>m>>n)
{
if(m==-1 && n==-1)
break;
for(int i=0; i<n; i++)
{
cin>>arr[i].J>>arr[i].F;
arr[i].ave = arr[i].J/arr[i].F;
}
sort(arr, arr+n, cmp);
double ret = 0.0;
for(int i=0; i<n; i++)
{
if(m >= arr[i].F)
{
ret += arr[i].J;
m -= arr[i].F;
}
else
{
ret += arr[i].J/arr[i].F*m;
break;///注意要break哦
}
}
printf("%.3lf\n",ret);
}
return 0;
}
