hdoj FatMouse' Trade 1009 （多重背包）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57207    Accepted Submission(s): 19174

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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500
 
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct zz{int v;int w;double c;}q[1010];int cmp(zz a,zz b){return a.c>b.c;}int main(){int n,m,i,j;while(scanf("%d%d",&m,&n),n!=-1&&m!=-1){for(i=1;i<=n;i++){scanf("%d%d",&q[i].v,&q[i].w);q[i].c=q[i].v*1.0/q[i].w;}sort(q+1,q+n+1,cmp);double sum=0;for(i=1;i<=n;i++){if(q[i].w<=m){sum+=q[i].v;m-=q[i].w;}else{sum+=m*q[i].c;break;}}printf("%.3lf\n",sum);}return 0;}