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
i pounds of JavaBeans and requires F
i
i pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J
i
i* a% pounds of JavaBeans if he pays F
i
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
i and F
i
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
#include<cstdio>
#include<algorithm>
using namespace std;
struct node{
int f;
int j;
double s;
}cat[10005];
bool cmp(node a,node b){
return a.s > b.s;
}
int main(){
int m,n;
while (scanf("%d%d",&m,&n),m != -1||n != -1){
for (int i = 0; i < n; ++i){
scanf("%d%d",&cat[i].j,&cat[i].f);
cat[i].s = cat[i].j*1.0/cat[i].f;
}
sort(cat,cat+n,cmp);
double sum = 0;
for (int i = 0; i < n; ++i){
if (m >= cat[i].f){
sum += cat[i].j;
m -= cat[i].f;
}
else {
sum += m*cat[i].j*1.0/cat[i].f;
break;
}
}
printf("%.3lf\n",sum);
}
return 0;
}
while(m){
if (m >= cat[i].f){
sum += cat[i].j;
m -= cat[i].f;
++i;
}
else {
sum += m*cat[i].j*1.0/cat[i].f;
m = 0;
}
这么写是不行的 = = wa了好久。。。注意 m剩余就死循环了 = =
哎。。
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