USACO: Beef McNuggets

Beef McNuggets
Hubert Chen

Farmer Brown's cows are up in arms, having heard that McDonalds is considering the introduction of a new product: Beef McNuggets. The cows are trying to find any possible way to put such a product in a negative light.

One strategy the cows are pursuing is that of `inferior packaging'. ``Look,'' say the cows, ``if you have Beef McNuggets in boxes of 3, 6, and 10, you can not satisfy a customer who wants 1, 2, 4, 5, 7, 8, 11, 14, or 17 McNuggets. Bad packaging: bad product.''

Help the cows. Given N (the number of packaging options, 1 <= N <= 10), and a set of N positive integers (1 <= i <= 256) that represent the number of nuggets in the various packages, output the largest number of nuggets that can not be purchased by buying nuggets in the given sizes. Print 0 if all possible purchases can be made or if there is no bound to the largest number.

The largest impossible number (if it exists) will be no larger than 2,000,000,000.

PROGRAM NAME: nuggets

INPUT FORMAT

Line 1:	N, the number of packaging options
Line 2..N+1:	The number of nuggets in one kind of box

SAMPLE INPUT (file nuggets.in)

3
3
6
10

OUTPUT FORMAT

The output file should contain a single line containing a single integer that represents the largest number of nuggets that can not be represented or 0 if all possible purchases can be made or if there is no bound to the largest number.

SAMPLE OUTPUT (file nuggets.out)

17

 

 
解题思路：

此题，orz。。。

想了很久，原来简单的DP就可以过了

重点在于这样一个结论：（假设存在最大的，无法得到的数），那么此数不大于最大的两个基础数的最小公倍数，从这个数据来说，就是255×256了。可见题目说"The largest impossible number (if it exists) will be no larger than 2,000,000,000. "是吓唬人的。

另个一结论是，所求的数无限大当且仅当所有基础数的最小公倍数不是1。这个很好理解。

/* ID: geterns1 PROG: nuggets LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; bool nums[100000]; int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b); } int main(){ freopen("nuggets.in", "r", stdin); freopen("nuggets.out", "w", stdout); // int Min = 300; int cnt, list[10]; scanf("%d", &cnt); for (int i = 0; i < cnt; i ++){ scanf("%d", list + i); if (list[i] < Min) Min = list[i]; } // int Gcd = list[0]; for (int i = 1; i < cnt; i ++) Gcd = gcd(Gcd, list[i]); if (Gcd ^ 1){ printf("0/n"); goto END; } // int count, last = 0, theNum = 0; memset(nums, 0, sizeof(nums)); nums[0] = true; int k; while (true){ count = 0; while (count ^ Min){ theNum ++; for (k = 0; k < cnt; k ++){ if (theNum - list[k] > -1 && nums[theNum - list[k]]){ nums[theNum] = true; count ++; break; } } if (k == cnt){ last = theNum; break; } } if (count == Min){ printf("%d/n", last); goto END; } } END: fclose(stdin); fclose(stdout); return 0; }