POJ 3111 K Best（最大化平均值）

题意：保留k个宝石，使得平均值尽可能的大。 具体跟上题一样。

解题思路：二分的方法跟上一题一样， 每次把排序的序列保存下来， 输出id就好了。

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

Memory: 3068 KB Time: 4407 MSLanguage: G++ Result: Accepted

#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<cctype>#include<list>#include<iostream>#include<map>#include<queue>#include<set>#include<stack>#include<vector>using namespace std;#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)int buf[10];inline long long read(){    long long x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return x*f;}inline void writenum(int i){    int p = 0;    if(i == 0) p++;    else while(i)        {            buf[p++] = i % 10;            i /= 10;        }    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);}/**************************************************************/#define MAX_N 100005const int INF = 0x3f3f3f3f;int v[MAX_N];int w[MAX_N];struct node{    double v;    int id;}c[MAX_N];bool cmp(node a, node b){    return a.v > b.v;}int n, k;bool f(double x){    for(int i = 0 ; i < n ; i++)    {        c[i].v = v[i] - x * w[i];        c[i].id = i;    }    sort(c, c + n, cmp);    double sum = 0;    for(int i = 0 ; i < k ; i++)    {        sum += c[i].v;    }    return sum >= 0;}int main(){    while(~scanf("%d%d", &n, &k))    {        for(int i = 0 ; i < n ; i++)        {            v[i] = read();            w[i] = read();        }        double lb = 0, ub = INF;        while(ub - lb > 1e-8)        {            double mid = (ub + lb) / 2;            if(f(mid)) lb = mid;            else ub = mid;        }        for(int i = 0 ; i < k - 1 ; i++)        {            cout<<c[i].id + 1<<" ";        }        cout<<c[k -1].id + 1<<endl;    }    return 0;}