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.
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
解题思路:
简单贪心算法 , 计算单价 , 按单价从大到小排个序 , 先挑大的拿。
代码:
#include<cstdio>
#include<algorithm>
using namespace std;
typedef struct{
int j; //可以换取的
int f; //用来换的
double v;
}value;
value ans[1010];
bool method(value a , value b){
return a.v > b.v;
}
int main(){
//freopen("D://testData//1009.txt" , "r" , stdin);
int m , n , i ;
double result;
while(scanf("%d %d",&m , &n) != EOF){
if(m == -1 && n == -1)
break;
result = 0;
for(i = 0 ; i < n ; i ++){
scanf("%d %d",&ans[i].j , &ans[i].f);
ans[i].v = (double)ans[i].j / ans[i].f;
}
sort(ans , ans + n , method);
for(i = 0 ; i < n ; i ++){
if(m >= ans[i].f){
result = result + ans[i].j;
m = m - ans[i].f;
}else{
result = result + ans[i].v * m;
break;
}
}
printf("%.3f\n",result);
}
return 0;
}