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 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.
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.
13.333
31.500
题解:贪心 从 $J[i] / F[i]$ 最大的开始选
时间复杂度:$O(N * logN)$
代码:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int N, M;
struct Node{
double j;
double f;
double pri;
}node[maxn];
bool cmp(const Node& a, const Node& b) {
return a.pri > b.pri;
}
int main() {
while(~scanf("%d%d", &M, &N)) {
if(N == -1 && M == -1) break;
for(int i = 1; i <= N; i ++) {
scanf("%lf%lf", &node[i].j, &node[i].f);
node[i].pri = node[i].j / node[i].f;
}
double num = 0.0;
sort(node + 1, node + 1 + N, cmp);
for(int i = 1; i <= N; i ++) {
if(M >= node[i].f) {
num += node[i].j;
M -= node[i].f;
}
else {
num += M * node[i].j / node[i].f;
break;
}
}
printf("%.3lf\n", num);
}
return 0;
}