HDUOJ 1009
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
據說是貪心之類的東西,但是是水題,沒涉及演算法,就是一種思想…
#include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> using namespace std; struct node { double a, b, c; }data[1005]; int cmp(node x, node y) { return x.c > y.c; } int main() { int n, m; while(scanf("%d%d", &n, &m) && (n != -1 && m != -1)) { for(int i = 0; i < m; i++) { scanf("%lf%lf", &data[i].a, &data[i].b); data[i].c = data[i].a / data[i].b; } sort(data, data + m, cmp); // for(int i = 0; i < m; i++) // { // cout << data[i].a << " " << data[i].b << " " << data[i].c << '\n'; // } double sum = 0; for(int i = 0; i < m; i++) { if(n <= data[i].b) { sum += n * data[i].c; // cout << "最後拿了" << n * data[i].c << '\n'; break; } else { n -= data[i].b; sum += data[i].a; // cout << "拿了" << data[i].a << '\n'; } } printf("%.3f\n", sum); } return 0; }