POJ 3111 K Best + poj 2976 Dropping tests (二分， 最大化平均值）

这两题类似， 都是用二分枚举 x    然后通过式子变形判断是否大于0

注意精读问题和循环次数， 过大会TLE

题目：

E - Dropping tests

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Appoint description:

Description

In a certain course, you take n tests. If you get a_i out of b_i questions correct on test i, your cumulative average is defined to be

.

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

Input

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

Sample Input

3 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

Sample Output

83
100

Hint

To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).

代码：

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;

#define LL long long
#define MAXN 1000+10
int n,k;
double a[MAXN]  ,b[MAXN];
double tmp[MAXN];
bool cmp(double x , double y)
{
    return x>y;
}
bool C(double mid)
{
    int  num = n-k;
//    cout<<"mid" <<mid<<endl;
    for(int i=0;i<n;i++)
    {
        tmp[i] = 100*a[i] - mid*b[i];
    }
    sort( tmp, tmp+n,cmp);

    double ret= 0;
    for(int i=0;i<num;i++)
    {
        ret += tmp[i];
      //  cout<<ret <<" " <<tmp[i]<<endl;
    }

    if( ret >= 0) return true;
    return false;
}


int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if( n==0 )break;

        for(int i=0;i<n;i++)
        {
            scanf("%lf",&a[i]);
        }

        for(int i=0;i<n;i++)
        {
            scanf("%lf",&b[i]);
        }

        double lb = 0 , ub =101;
        double mid;
        LL ans = 0;
        for(int i=0;i<100;i++)
        {
            double mid = (lb+ub)/2;
            if(C(mid))
            {
                lb =mid;
                ans = round(mid);
            }
            else
                ub = mid;
        }
        printf("%lld
",ans);

    }

    return 0;
}

题目：

F - K Best

Time Limit:8000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Appoint description:

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 2
1 1
1 2
1 3

Sample Output

1 2

代码：

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
#define MAXN 100000+10
#define LL long long
int v[MAXN];
int w[MAXN];
int ans[MAXN];
struct P
{
    double t;
    int pos;
    bool operator < (const P& x) const
    {
        return t>x.t;
    }
};
P tmp[MAXN];
int n,k;
bool C(double mid)
{
    for(int i=0;i<n;i++)
    {
        tmp[i].t = v[i] - mid*w[i];
        tmp[i].pos=i;
    }
    sort(tmp,tmp+n);
    double ret = 0;
    for(int i=0;i<k;i++)
    {
        ret += tmp[i].t;
        ans[i] = tmp[i].pos;
    }
    return ret>=0;
}

int main()
{
    scanf("%d%d",&n,&k);

    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&v[i],&w[i]);
    }
    double lb = 0 , rb = 10000000;

    for(int i=0;i<50;i++)
    {
        double mid = (lb+rb)/2;
        if( C(mid) )lb = mid;
        else
            rb = mid;
    }
    for(int i=0;i<k;i++)
    {
        if(i != k-1)
            printf("%d ",ans[i]+1);
        else
            printf("%d
",ans[i]+1);
    }
    return 0;
}