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hdu1009
FatMouse‘ Trade |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 4555 Accepted Submission(s): 1354 |
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 |
刚开始用了一个比较普通的算法：1循环输入数据，2根据“好坏”排序数据，3得到最优解；验证数据正确，可能是这个算法太繁琐导致超时，遗憾不能ac
1 #include<stdio.h> 2 int main() 3 { 4 int m,n,i,s; 5 double j[1005],f[1005],t[1005],tmpt,tmpj,tmpf,cha,max=0; 6 while(scanf("%d%d",&m,&n)!=EOF) 7 { 8 if(m<=0&&n<=0) 9 break;10 max=0;11 for(i=0;i<n;i++){12 scanf("%lf%lf",&j[i],&f[i]);13 t[i]=j[i]/f[i];14 }15 for(i=0;i<n-1;i++)16 for(s=1;s<n;s++)17 if(t[i]<t[s])18 {19 tmpt=t[i]; tmpj=j[i]; tmpf=f[i];20 t[i]=t[s]; j[i]=j[s]; f[i]=f[s];21 t[s]=tmpt; j[s]=tmpj; f[s]=tmpf;22 }23 24 i=0;25 while(m)26 {27 if(m>=f[i]){28 max +=j[i];29 m -=f[i];30 i++;31 }32 else{33 cha=m/f[i];34 max +=j[i]*cha;35 m=0;36 i++;37 }38 if(i>=n)39 break;40 }41 printf("%.3f\n",max);42 i=0;max=0;43 44 }45 return 0;46 }
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