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HDU1009 贪心思想
FatMouse‘ Trade
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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
这个题的大意是:它有M猫食,N个房间,每个房间有f[i]猫食,j[i]量的 javabean,按一定的比例拿猫食来保护javabean。问一共有M猫食最多能保护多少javabean。
简单的贪心,只需按每个房间的javabean和猫食的比例从高到底排序,然后贪心即可。
代码中我处理了输入时f[i]等于0的情况,但交上去WA,我删去之后就AC了,看来没必要处理啊,,
#include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> using namespace std; struct food { int f,j; double ave; }s[1010]; int cmp(const food &a,const food &b) { return a.ave>b.ave; } int main() { int i,k,n,m; double ans; while(scanf("%d%d",&m,&n)) { if(m==-1 && n==-1) break; for(i=0;i<n;i++) { scanf("%d%d",&s[i].j,&s[i].f); s[i].ave=(double)s[i].j/s[i].f; } sort(s,s+n,cmp); ans=0; for(i=0;i<n;i++) { if(s[i].f<=m) { ans+=(double)s[i].j; m-=s[i].f; } else { ans+=s[i].ave*m; break; } } printf("%.3f\n",ans); } return 0; }
HDU1009 贪心思想
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