POJ 3111 K Best 【二分：最大化平均值】

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 10371 Accepted: 2672Case Time Limit: 2000MS Special Judge

Description

Demy has n jewels. Each of her jewels has some value v_i and weight w_i.

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, …, i_k} as

.

Demy would like to select such k jewels that their specific value is maximal possible. Help her to do so.

Input

The first line of the input file contains n — the number of jewels Demy got, and k — the number of jewels she would like to keep (1 ≤ k ≤ n ≤ 100 000).

The following n lines contain two integer numbers each — v_i and w_i (0 ≤ v_i ≤ 10⁶, 1 ≤ w_i ≤ 10⁶, both the sum of all v_i and the sum of all w_i do not exceed 10⁷).

Output

Output k numbers — the numbers of jewels Demy must keep. If there are several solutions, output any one.

Sample Input

3 21 11 21 3

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

#include<iostream>  #include<cstdio>  #include<cstring>  #include<string>  #include<cmath>  #include<queue>  #include<stack>  #include<vector>  #include<map>  #include<algorithm>  using namespace std;  #define ll long long  #define ms(a,b)  memset(a,b,sizeof(a))  const int M=1e5+10;  const int inf=0x3f3f3f3f;  const int mod=1e9+7; int i,j,k,n,m; int a[M]; int b[M];double y[M];struct node{int num;int v;int w;double y;bool operator < (const node&x) const    {        return y>x.y;    }}p[M];bool cmp(node a,node b){ return a.y>b.y;}int kk[M];bool check(double mid){for(int i=1;i<=n;i++)p[i].y=p[i].v-p[i].w*mid;sort(p+1,p+n+1);double sum=0;for(int i=1;i<=k;i++){       sum+=p[i].y;   kk[i]=p[i].num;}return sum>=0;}int main(){while(~scanf("%d%d",&n,&k)){if(n==0&&k==0)break;for(int i=1;i<=n;i++){scanf("%d %d",&p[i].v,&p[i].w);    p[i].num=i;     }        double lb=0,ub=1000000000;    for(int i=0;i<100;i++){ double mid=(lb+ub)/2;         if(check(mid))lb=mid; else ub=mid;     }     for(int i=1;i<=k;i++)if(i>1)printf(" %d",kk[i]);else printf("%d",kk[i]); printf("\n");   }   return 0;}