做一个机智的胖老鼠（贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 52802 Accepted Submission(s): 17611



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

Author

CHEN, Yue

机智的胖老鼠，类似于背包问题，没什么好说的，但是代码WA了半天，原因就是一个应该是double的变量定义成了Int。

int main()
{
double M, N;
int i;
double max;

while(1)
{
scanf("%lf %lf", &M, &N);//输入猫粮数和房间数。
if(M == -1 || N == -1)   //跳出条件
break;
double result = 0;
double F[sz], J[sz], R[sz];//猫粮，Javabean，及其比例
int ii;
for(i = 0; i < N; i ++)<span style="white-space:pre">	</span>
{
scanf("%lf %lf", &J[i], &F[i]);
R[i] = J[i]*1.0 / F[i];
}
while(M && result != N) //跳出条件：猫粮用完或者Javabean全部get
{
max = 0;
for(i = 0; i < N; i ++)<span style="white-space:pre">	</span>//每一次都找最划算的房间
{
if(R[i] > max)
{
max = R[i];<span style="white-space:pre">	</span>//记录最大值，用于比较
ii = i;<span style="white-space:pre">		</span>//记录该房间的号码，用于交易
}
}
if(M > F[ii])<span style="white-space:pre">		</span>//足够提供这个房间所需的全部猫粮
{
M -= F[ii];
result += J[ii];
R[ii] = 0;<span style="white-space:pre">	</span>//交换完成置零，不再参与
}
else<span style="white-space:pre">			</span>//不够，用完就跳出了
{
result += (J[ii]*M*1.0)/(F[ii]*1.0);
M = 0;
R[ii] = 0;
}
}
printf("%.3f\n",result);