Codeforces Round #311(Div.2) (2/5)
A Ilya and Diplomas
Soon a school Olympiad in Informatics will be held in Berland, n schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the n participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least min1 and at most max1 diplomas of the first degree, at leastmin2 and at most max2 diplomas of the second degree, and at least min3 and at most max3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all n participants of the Olympiad will receive a diploma of some degree.
Input
The first line of the input contains a single integer n (3 ≤ n ≤ 3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers min1 and max1 (1 ≤ min1 ≤ max1 ≤ 106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers min2 and max2 (1 ≤ min2 ≤ max2 ≤ 106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers min3 and max3 (1 ≤ min3 ≤ max3 ≤ 106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that min1 + min2 + min3 ≤ n ≤ max1 + max2 + max3.
Output
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
Sample test(s)
input
6
1 5
2 6
3 7
output
1 2 3
input
10
1 2
1 3
1 5
output
2 3 5
input
6
1 3
2 2
2 2
output
2 2 2
题目类型：贪心
算法分析：首先要保证每个证书的人数要大于等于限制最小值，然后贪心地选取级别高的证书
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 |
#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; const int INF = 0x7FFFFFFF; const int MOD = 1000000000 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 100000 + 66; int main() { // ifstream cin ("aaa.txt"); int n, a1, a2, b1, b2, c1, c2; cin >> n; cin >> a1 >> a2 >> b1 >> b2 >> c1 >> c2; int aa = a1, bb = b1, cc = c1; while (aa + bb + cc < n) { if (aa < a2) aa++; else if (bb < b2) bb++; else if (cc < c2) cc++; } cout << aa << " " << bb << " " << cc << endl; return 0; } |
B Pasha and Tea
Pasha decided to invite his friends to a tea party. For that occasion, he has a large teapot with the capacity of wmilliliters and 2n tea cups, each cup is for one of Pasha's friends. The i-th cup can hold at most ai milliliters of water.
It turned out that among Pasha's friends there are exactly n boys and exactly n girls and all of them are going to come to the tea party. To please everyone, Pasha decided to pour the water for the tea as follows:
- Pasha can boil the teapot exactly once by pouring there at most wmilliliters of water;
- Pasha pours the same amount of water to each girl;
- Pasha pours the same amount of water to each boy;
- if each girl gets xmilliliters of water, then each boy gets 2x milliliters of water.
In the other words, each boy should get two times more water than each girl does.
Pasha is very kind and polite, so he wants to maximize the total amount of the water that he pours to his friends. Your task is to help him and determine the optimum distribution of cups between Pasha's friends.
Input
The first line of the input contains two integers, n and w (1 ≤ n ≤ 105, 1 ≤ w ≤ 109) — the number of Pasha's friends that are boys (equal to the number of Pasha's friends that are girls) and the capacity of Pasha's teapot in milliliters.
The second line of the input contains the sequence of integers ai (1 ≤ ai ≤ 109, 1 ≤ i ≤ 2n) — the capacities of Pasha's tea cups in milliliters.
Output
Print a single real number — the maximum total amount of water in milliliters that Pasha can pour to his friends without violating the given conditions. Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 6.
Sample test(s)
input
2 4
1 1 1 1
output
3
input
3 18
4 4 4 2 2 2
output
18
input
1 5
2 3
output
4.5
Note
Pasha also has candies that he is going to give to girls but that is another task...
题目类型：贪心
算法分析：将所有杯子的容积从小到大排序，然后找到前n个体积和后n个体积的最小值a、b，以a和b/2中的相对较小的作为每个x，最后直接计算即可，注意输出时要使用控制精度的语句，否则会丢失精度！！！
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 |
#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 66; int a[maxn*2]; int main() { // CPPFF; int n, w; double res = 0; cin >> n >> w; for (int i = 1;i <= 2 * n; i++) cin >> a[i]; sort (a + 1, a + 1 + 2 * n); double minvala = a[1], minvalb = a[n+1]; double temp; temp = minvalb / 2.0; temp = min (minvala, temp); if (w <= 3 * n * temp + EPS) cout << w << endl; else cout << fixed << setprecision(6) << 3 * n * temp + EPS << endl; return 0; } |