最大化平均值

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 2
1 1
1 2
1 3

Sample Output

1 2


题意 ： 从 n 个物品中选出 k 个，问怎么选平均总价值最大 ？

思路 ：

　　先取前k个元素算出S0 =∑(vi/wi) 作为初始值 ，然后对每一个元素(n个)求s=vi-x*wi

　　要体会二分搜索过程的关键点，二分的 mid 值就是你所要一直求得值。

代码 ：

const int eps = 1e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long

int n, k;
struct node
{
    int v, w, pt;
    double s;
    
    bool operator< (const node& ff)const{
        return s > ff.s;
    }
}pre[eps];

bool check(double x){
    for(int i = 1; i <= n; i++){
        pre[i].s = 1.0*pre[i].v - x*pre[i].w;
    }    
    sort(pre+1, pre+n+1);
    double sum = 0;
    for(int i = 1; i <= k; i++){
        sum += pre[i].s;
    }
    return sum >= 0;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", sttout);
    
    while(~scanf("%d%d", &n, &k)){
        for(int i = 1; i <= n; i++){
            pre[i].pt = i;
            scanf("%d%d", &pre[i].v, &pre[i].w);
        }
        double l = 0, r = 1.0*inf;
        
        for(int i = 1; i <= 200; i++){
            double mid = (l + r) / 2.0;
            if (check(mid)) l = mid;
            else r = mid;
        }
        
        for(int i = 1; i < k; i++){
            printf("%d ", pre[i].pt);
        }
        printf("%d
", pre[k].pt);
        //printf("%.2f
", ans);
    }
    return 0;
}