航院1009： FatMouse’s Trade
航院1009 FatMouse’s Trade 解题报告
Problem Description
FatMouse prepared M pounds ofcat food, ready to trade with the cats guarding the warehouse containing hisfavorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans andrequires F[i] pounds of cat food. FatMouse does not have to trade for all theJavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if hepays F[i]* a% pounds of cat
food. Here a is a real number. Now he is assigningthis homework to you: tell him the maximum amount of JavaBeans he can obtain.
Input
The input consists of multipletest cases. Each test case begins with a line containing two non-negativeintegers M and N. Then N lines follow, each contains two non-negative integersJ[i] and F[i] respectively. The last test case is followed by two -1's. Allintegers are not greater than 1000.
Output
For each test case, print in asingle line a real number accurate up to 3 decimal places, which is the maximumamount 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：
1.先计算出每一个房间平均一单位猫粮可以交换多少豆子。
2.将平均一单位猫粮交换的豆子的数量由大到小进行排序。
3.当总猫粮数大于当前猫粮数时，累加这一房间的豆子，同时总猫粮数减去当前房间猫粮数。
4.当总猫粮数小于等于当前猫粮数时，用剩余猫粮数乘以此房间平均一单位猫粮可以交换的豆子。
5.输出豆子数。
错误原因：
1. 未充分考虑各种情况。题目中说每个数据都是非负整数，未考虑M大于F[i]的和的情况。
2. 进行排序的时候，未将就j[i]和f[i]分别进行排序。
经验总结：
1. 如果程序执行到固定某一步时结束，则可以使用完整循环，在需要结束时判断并跳出即可。
正确程序代码：
#include<stdio.h>
double j[1005]={0},f[1005]={0};
int main()
{
int n, i, k;
double m,t1, t2;
double sum;
while(scanf("%lf%d",&m, &n)!=EOF)
{
if(m==-1&& n==-1)
{
return0;
}
sum=0.0;
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&j[i], &f[i]);
}
for(i=1;i<=n-1;i++)
{
for(k=i+1;k<=n;k++)
{
if(j[i]/f[i]<j[k]/f[k])
{
t1=j[i];
j[i]=j[k];
j[k]=t1;
t2=f[i];
f[i]=f[k];
f[k]=t2;
}
}
}
for(i=1;i<=n;i++)
{
if(m>f[i])
{
sum+=j[i];
m=m-f[i];
}
else
{
sum+=(double)m*j[i]/f[i];
break;
}
}
printf("%.3lf\n", sum);
}
return 0;
}
