/* 摘抄(简单贪心) FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33765 Accepted Submission(s): 10999 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 ZJCPC2004 Recommend JGShining 简单贪心 */ #include <iostream> //很简单的一道贪心，不知道为什么会错那么多次，搞不懂 #include <algorithm> #include <cstdio> #include <string.h> #include <queue> #include <stdlib.h> using namespace std; typedef struct node { int j; int f; double d; }MC; MC a[1003]; bool cmp(MC a,MC b) { return a.d>b.d; } int main() { int n,m,i; double t; while(scanf("%d%d",&n,&m)) { if(n==-1&&m==-1) break; for(i=0;i<m;i++) { scanf("%d%d",&a[i].j,&a[i].f); a[i].d=1.0*a[i].j/a[i].f; } sort(a,a+m,cmp); t=0; for(i=0;i<m;i++) { if(n>a[i].f) { t+=a[i].j; n-=a[i].f; } else {t+=n*a[i].d;break;}//老是在这出问题、郁闷---现在明白了，有当m=0时的情况， //所以这个不能写在循环外面。 } printf("%.3lf ",t); } return 0;//现在明白了，此题的变态之处在于，m和f可以为0，伤人呀！sort居然可以排序分母为零的，强悍！！！ }