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
FatMouse准备了M磅的猫食,准备与守卫仓库的猫交易,这些猫包含他最喜欢的食物,JavaBean。
仓库有N个房间。第i间房间包含J [I]磅的JavaBeans,并且需要F [i]磅的猫粮。 FatMouse不必交易房间内的所有JavaBeans,相反,如果他付给F [i] * 1磅的猫粮,他可能会得到1磅的JavaBeans。这里是一个实数。现在他正在为你分配这个作业:告诉他他可以获得的最大JavaBeans数量。
输入
输入由多个测试用例组成。每个测试用例都以包含两个非负整数M和N的行开始。然后N行包含两个非负整数J [i]和F [i]。最后一个测试用例后面跟着两个-1。所有整数不大于1000。
产量
对于每个测试用例,在一行中打印一个真实数字,精确到3位小数,这是FatMouse可以获得的最大JavaBean数量。
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
解题思路:贪心思想,算出单价排序,然后从大到小去取.看代码
#include <iostream> #include <math.h> #include <stdio.h> #include <algorithm> using namespace std; struct stu { double ja,mao,dj; } c[1000]; bool cmp(stu x,stu y) { return x.dj>y.dj; } int main() { int m,n; while(scanf("%d %d",&m,&n),m!=-1&&n!=-1) { double a[5000],b[5000],ans=0; for(int i=1; i<=n; i++) { scanf("%lf %lf",&c[i].ja,&c[i].mao); c[i].dj=c[i].ja/c[i].mao; } sort(c+1,c+n+1,cmp); for(int i=1;i<=n;i++) { if(m>=c[i].mao) { m-=c[i].mao; ans+=c[i].ja; } else { ans+=m*c[i].dj; break; } } printf("%.3lf ",ans); } return 0; }