POJ 2976 Dropping tests、3111 K Best （二分搜索）
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7327 | Accepted: 2548 |
Description
In a certain course, you take n tests. If you get ai out of bi 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 ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 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;
}
| 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 vi and weight wi.
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, …, ik} 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 — vi and wi (0 ≤ vi ≤ 106, 1 ≤ wi ≤ 106, both the sum of all vi and the sum of all wi 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;
}