POJ 2976 Dropping tests、3111 K Best （二分搜索）

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7327		Accepted: 2548

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

把式子展开然后就发现是个贪心了 - - Orz

两道一样的题 就贴下代码好了

AC代码如下：

//
//  Created by TaoSama on 2015-04-28
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n, k, a[1005], b[1005];
double y[1005];

bool check(double x) {
    for(int i = 1; i <= n; ++i)
        y[i] = a[i] - x * b[i];
    sort(y + 1, y + 1 + n);

    /*for(int i = 1; i <= n; ++i)
        cout<<y[i]<<' ';
    cout<<endl<<endl; */

    double sum = 0;
    for(int i = 1; i <= n - k; ++i)
        sum += y[n - i + 1];
    return sum >= 0;
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(cin >> n >> k && (n + k)) {
        for(int i = 1; i <= n; ++i) cin >> a[i];
        for(int i = 1; i <= n; ++i) cin >> b[i];

        double l = 0, r = 1;
        for(int i = 0; i < 100; ++i) {
            double mid = (l + r) / 2;
            if(check(mid)) l = mid;
            else r = mid;
        }
        cout << (int)(l * 100 + 0.5) << '\n';
    }
    return 0;
}

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 6638		Accepted: 1753
Case 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 2
1 1
1 2
1 3

Sample Output

1 2

好像100次T了 - - 然后改成了50次 Orz

AC代码如下：

//
//  Created by TaoSama on 2015-04-28
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n, k, v[N], w[N];
struct Jewel {
    int id;
    double y;
    bool operator<(const Jewel& rhs) const {
        return y > rhs.y;
    }
} a[N];

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

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) scanf("%d%d", v + i, w + i);

    double l = 0, r = 1e7 + 1;
    for(int i = 0; i < 50; ++i) {
        double mid = (l + r) / 2;
        if(check(mid)) l = mid;
        else r = mid;
    }

    for(int i = 1; i <= k; ++i)
        printf("%d%c", a[i].id, i == k ? '\n' : ' ');
    return 0;
}