- FatMouse' Trade
- Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
- Total Submission(s): 103515 Accepted Submission(s): 36159
- 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
题意:
这一道题意思就是老鼠用猫食物换取自己最喜爱的食物 javaBean 的过程, 当然换取的最终结果是保证最后的 JavaBean 是最多的,
而且是当自己手中的猫食物小于每个仓库所需交换的猫食物时候, 可以手中有多少就交换多少.
所以在解这道题时候要想到按照每个仓库 javaBean 最大的比率排序才能保证最后的交换的 javaBean 是最大的.
- #include<iostream>
- #include<cstdio>
- #include<algorithm>
- using namespace std;
- const int maxn = 1e3+10;
- int m,n;
- struct node{
- int j,f;
- double r;
- } food[maxn];
- class cmp{
- public:
- bool operator()(node a,node b)const{
- return a.r>b.r;
- }
- };
- int main(){
- while(~scanf("%d%d",&m,&n)){
- if(m==-1) break;
- for(int i=0; i<n; i++ ){
- cin>>food[i].j>>food[i].f;
- food[i].r=1.0*food[i].j/food[i].f;
- }
- sort(food,food+n,cmp());
- double ans=0;
- for( int i=0; i<n; i++ ){
- if(m>=food[i].f){
- ans+=food[i].j;
- m-=food[i].f;
- }
- else{
- if(m==0) break;
- else{
- ans+=m*food[i].r;
- break;
- }
- }
- }
- printf("%.3f\n",ans);
- }
- return 0;
- }
来源: http://www.bubuko.com/infodetail-2985544.html