FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 77314 Accepted Submission(s): 26527
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 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
题本身并不难,但是要多考虑下,如,当小鼠没有猫粮,或者有些豆子白送,或者没有交换豆子的房间,或者小鼠没猫粮但是刚好只有一间屋子,且是白送猫粮的情况,总之多多考虑
#include<iostream> #include<cstdio> using namespace std; int main() { double m,n; while(cin>>m>>n) { double a[1001][3]={0},cnt=0; if(m==(-1)&&n==(-1)) break; if(n==0) {printf("%.3f\n",cnt);continue;} int i,j,k; double t,f=1; for(i=1;i<=n;++i) { scanf("%lf%lf",&a[i][1],&a[i][2]); } for(i=1;i<=n;++i) { if(a[i][2]==0) {cnt+=a[i][1];continue;} else {k=i; for(j=i+1;j<=n;++j) { if(a[j][2]==0) continue; else if((a[j][1]/a[j][2])>(a[k][1]/a[k][2])) k=j; } if(k!=i) { t=a[i][1],a[i][1]=a[k][1],a[k][1]=t; t=a[i][2],a[i][2]=a[k][2],a[k][2]=t; } } } for(i=1;i<=n;++i) { if(a[i][2]==0) continue; else { if(m<a[i][2]) {printf("%.3f\n",cnt+(m/a[i][2])*a[i][1]);m=0;f=0;break;} else if(m==a[i][2]) {printf("%.3f\n",cnt+a[i][1]);m=0;f=0;break;} else { m-=a[i][2]; cnt+=a[i][1]; f=1; } } } if(f==1) printf("%.3f\n",cnt); } return 0; }