FatMouse' Trade |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 4921 Accepted Submission(s): 1476 |
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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. |
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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.
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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.
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Sample Input
5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1 |
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Sample Output
13.333 31.500 |
#include <stdio.h>
#include <stdlib.h>
int main()
{
int M, N;
while (scanf("%d%d", &M, &N) != EOF)
{
int i, j, J[1000], F[1000], temp1;
double ratio[1000], JavaBeans = 0, temp;
if ((M == -1) && (N == -1))
{
return 0;
}
for (i = 0; i<N; i++)
{
scanf("%d%d", &J[i], &F[i]);
ratio[i] = (J[i] * 1.0) / F[i];
}
for (i = 0; i<N; i++)
{
for (j = 0; j<N - i - 1; j++)
{
if (ratio[j]<ratio[j + 1])
{
temp = ratio[j];
ratio[j] = ratio[j + 1];
ratio[j + 1] = temp;
temp1 = F[j];
F[j] = F[j + 1];
F[j + 1] = temp1;
temp1 = J[j];
J[j] = J[j + 1];
J[j + 1] = temp1;
}
}
}
for (i = 0; i<N; i++)
{
if (M >= F[i])
{
JavaBeans = J[i] + JavaBeans;
M = M - F[i];
}
else
{
JavaBeans = JavaBeans + ratio[i] * M;
break;
}
}
printf("%.3lf\n", JavaBeans);
}
return 0;
}