Painter
The local toy store sells small fingerpainting kits with between three and twelve 50ml bottles of paint, each a different color. The paints are bright and fun to work with, and have the useful property that if you mix X ml each of any three different colors, you get X ml of gray. (The paints are thick and "airy", almost like cake frosting, and when you mix them together the volume doesn't increase, the paint just gets more dense.) None of the individual colors are gray; the only way to get gray is by mixing exactly three distinct colors, but it doesn't matter which three. Your friend Emily is an elementary school teacher and every Friday she does a fingerpainting project with her class. Given the number of different colors needed, the amount of each color, and the amount of gray, your job is to calculate the number of kits needed for her class.
Input
The input consists of one or more test cases, followed by a line containing only zero that signals the end of the input. Each test case consists of a single line of five or more integers, which are separated by a space. The first integer N is the number of different colors (3 <= N <= 12). Following that are N different nonnegative integers, each at most 1,000, that specify the amount of each color needed. Last is a nonnegative integer G <= 1,000 that specifies the amount of gray needed. All quantities are in ml.
Output
For each test case, output the smallest number of fingerpainting kits sufficient to provide the required amounts of all the colors and gray. Note that all grays are considered equal, so in order to find the minimum number of kits for a test case you may need to make grays using different combinations of three distinct colors.
Sample Input
3 40 95 21 0
7 25 60 400 250 0 60 0 500
4 90 95 75 95 10
4 90 95 75 95 11
5 0 0 0 0 0 333
0
Sample Output
2
8
2
3
4
Source
题目类型：贪心
算法分析：首先将读入的颜料按照体积的大小递增排序，然后得到初始的颜料套装数，并计算出使用初始套装后，每种颜料还剩的体积。接着每次取体积最大的三种颜料1毫升配成灰色染料，直到灰色染料已经配完或者是再使用一个颜料套装。注意这里只能每次枚举1毫升，否则会WA
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#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 = 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 600 + 66; int ans[maxn], g, temp[maxn]; int main() { // ifstream cin ("aaa.txt"); // freopen ("aaa.txt", "r", stdin); int n; while (cin >> n) { if (n == 0) break; for (int i = 0; i < n; i++) cin >> ans[i]; cin >> g; sort (ans, ans + n); int sum; if (ans[n-1] % 50 == 0) sum = ans[n-1] / 50; else sum = ans[n-1] / 50 + 1; for (int i = 0; i < n; i++) temp[i] = sum * 50 - ans[i]; while (g > 0) { sort (temp, temp + n, greater <int> () ); if (temp[2] <= 0) { sum++; for (int i = 0; i < n; i++) temp[i] += 50; } temp[0]--, temp[1]--, temp[2]--; g--; } cout << sum << endl; } return 0; } |
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