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

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal 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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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: 65536KTotal Submissions: 6638 Accepted: 1753Case 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

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;}