Description：
Demy has n jewels. Each of her jewels has some value vi and weight wi.
Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k best jewels for herself. She decided to keep such jewels that their specific value is as large as possible. That is, denote the specific value of some set of jewels S = {i1, i2, …, ik} as
[ s(S) = frac{sum_{j=1}^k v_{ij}}{sum_{j=1}^kw_{ij}}$$.
Demy would like to select such k jewels that their specific value is maximal possible. Help her to do so.
## Analysis：
不妨令
$$ frac{sum_{j=1}^k v_{ij}}{sum_{j=1}^kw_{ij}} geq x ]
则
[sum_{j=1}^k v_{ij} geq x sum_{j=1}^kw_{ij}
]
即
[sum_{j=1}^k v_{ij} - x sum_{j=1}^kw_{ij} geq 0
]
只需要判断二分的每个k是否满足 ≥ 0即可。
Code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define N 100001
#define INF 0x3f3f3f3f
using namespace std;
const double EPS = 0.00001;//精度
int v[N],w[N],ans[N],n,k;
struct Node{
int id;
double val;
bool operator < (const Node& x) const {
return val > x.val;
}
}f[N];
bool C(double x){
for(int i = 0;i < n;++i){
f[i].val = v[i] - w[i]*x;
f[i].id = i + 1;
}
sort(f,f + n);
double sum = 0;
for(int i = 0;i < k;++i){
sum += f[i].val;
ans[i] = f[i].id;
}
return sum >= 0;
}
void solve(){
double lb = 0,ub = INF;
while(lb + EPS < ub){
double mid = (lb + ub)/2;
if(C(mid)) lb = mid;
else ub = mid;
}
for(int i = 0;i < k;++i){
printf("%d ",f[i].id);
}
}
int main(){
scanf("%d%d",&n,&k);
for(int i = 0;i < n;++i){
scanf("%d %d",&v[i],&w[i]);
}
solve();
return 0;
}