杭电1009--FatMouse' Trade
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这是一道简单的贪心题,题意为Fatmouse想要尽可能多得守护自己喜欢的JavaBean,所以,i房间得用F[i]d的cat food换得守护J[i]的JavaBean,当JavaBean不足以换的所有时,则根据所有cat food占所需cat food的比例来乘以i房间JavaBean得到实际能够得到的JavaBean。
思路很简单,要用一定的cat food守护尽可能多得JavaBean,必须要房间的JavaBean多并且需要的cat food 少。所以这种涉及两方面因素的贪心,可以通过两方面的比率来判断优先取哪一个。此题应当优先取J[i]/F[i]大的。
Total Submission(s): 41918 Accepted Submission(s): 13938
思路很简单,要用一定的cat food守护尽可能多得JavaBean,必须要房间的JavaBean多并且需要的cat food 少。所以这种涉及两方面因素的贪心,可以通过两方面的比率来判断优先取哪一个。此题应当优先取J[i]/F[i]大的。
FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41918 Accepted Submission(s): 13938
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.
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
Author
CHEN, Yue
Source
ZJCPC2004
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#include<stdio.h>#include<algorithm>using namespace std;struct cat{ int j;//javabean int f;//cat fod double t;}B[1005];typedef struct cat cat;bool cmp(cat a,cat b){ return a.t>b.t;}int main(){ int m,n; while(scanf("%d%d",&m,&n)!=EOF) { int i; double sum=0; int k=0; int xx,yy; if(m==-1&&n==-1) break; for(i=0;i<n;i++) { scanf("%d%d",&xx,&yy); if(yy==0) sum+=xx; else { B[k].j=xx;//javabean B[k].f=yy;//cat food(搞清字母含义,不然混乱会加错,我交了两遍都是错在字母搞混含义导致出错) B[k].t=(double)xx/yy; k++; } }//考虑特殊的当房间不需要cat food时,不需要cat food sort(B,B+k,cmp);//进行优先排序 //需要考虑几种情况,当n=0,时,当m=0时m和n都不为0时。实际上n=0和m不为0并且n不为0可以归为一类 if(m==0) printf("%.3lf\n",sum);//直接输出不需要cat food的JavaBean总和 else { int temp=m; for(i=0;i<k;i++) { if(temp<=B[i].f) sum+=B[i].t*(double)temp; else sum+=(double)B[i].j; temp-=B[i].f; if(temp<=0) break; } printf("%.3lf\n",sum); }//n=0和m!=0&&n!=0的情况归为一类 } return 0;}
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