HDU 1009 FatMouse' Trade

原題傳送門：http://acm.hdu.edu.cn/showproblem.php?pid=1009

FatMouse' Trade

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

Total Submission(s): 42086    Accepted Submission(s): 14004

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  

Source

ZJCPC2004

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JGShining   |   We have carefully selected several similar problems for you:   1008  1050  1005  1051  1004 分析：這是一個簡單的貪心算法，部分背包問題，用結構體數組，以J/F比值升序排序即可。然後周遊數組的時候判斷每個房間能不能全拿，如果能就left-F,total+=J，不能就按照比例來取，在不能的時候Mouse用盡M的時候，這時候break就ok了，注意百分比計算的時候的強制轉換。  

//部分背包問題 貪心算法 HDU1009

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN = 1100;
struct Value{
    int J;
    int F;
    double j_f;
};
bool cmp(Value a,Value b){
    return a.j_f>b.j_f;
}
int N,M,j,f,leftM;
double total;
Value p[MAXN];
int main(){
   while(scanf("%d%d",&M,&N) ==2){
        if(M == -1&&N ==-1)break;
        leftM = M;
        total = 0;
         for(int i=0;i<N;i++){
            scanf("%d%d",&j,&f);
            p[i].J = j;
            p[i].F = f;
            p[i].j_f = ((double)j)/f;
         }
         sort(p,p+N,cmp);
         for(int i=0;i<N;i++){
            if(p[i].F<=leftM){
                leftM = leftM - p[i].F;
                total += p[i].J;
            }else{
                total += (((double)leftM)/p[i].F * p[i].J);
                break;
            }
         }
         printf("%.3f\n",total);
    }
    return 0;
}