CDZSC_2015寒假新人(1)——基础 c
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.
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
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1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 int main() 7 { 8 int m,n; 9 int f[1200],j[1200]; 10 double a[1200],b[1200]; 11 int next[1000]; 12 while((scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))) 13 { 14 for(int i=0;i<n;i++) 15 { 16 scanf("%d%d",&f[i],&j[i]); 17 a[i]=1.0*f[i]/j[i]; 18 } 19 double num=0; 20 for(int i=0;i<n;i++) 21 { 22 for(int k=i+1;k<n;k++) 23 { 24 if(a[i]<a[k]) 25 { 26 swap(a[i],a[k]); 27 swap(f[i],f[k]); 28 swap(j[i],j[k]); 29 } 30 } 31 if(j[i]<m) 32 { 33 num+=f[i]; 34 m-=j[i]; 35 } 36 else 37 { 38 39 num+=1.0*f[i]*m/j[i]; 40 m=0; 41 } 42 if(m==0) 43 { 44 break; 45 } 46 } 47 printf("%.3lf\n",num); 48 } 49 }