FatMouse' Trade
题目描述
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
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
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.
示例输入
5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
示例输出
13.333 31.500
提示
#include<stdio.h>
struct bean{
int contain, demand;
double bi;
}a[1000], b;
void quick_sort(bean s[], int l, int r){
if(l < r){
int i=l, j=r;
double x=s[l].bi;
b = s[l];
while(i < j){
while(i < j && s[j].bi >= x) j--;
if(i < j) s[i++] = s[j];
while(i < j && s[i].bi < x) i++;
if(i < j) s[j--] = s[i];
}
s[i] = b;
quick_sort(s, l, i-1);
quick_sort(s, i+1, r);
}
}
int main(){
int m, n;
while(scanf("%d%d", &m, &n) && m!= -1 && n!= -1){
int i;
for(i=0; i<n; i++){
scanf("%d%d", &a[i].contain, &a[i].demand);
a[i].bi = a[i].contain *1.0 / a[i].demand;
}
quick_sort(a, 0, n-1);
i = n-1;
double amount = 0;
while(m>0 && i>=0) {
if(m - a[i].demand >= 0) {
amount += a[i].contain;
m -= a[i].demand;
i--;
}
else {
amount += m * a[i].bi;
break;
}
}
printf("%.3lf
", amount);
}
return 0;
}