Canonical Coin Systems

题目描述

A coin system S is a finite (nonempty) set of distinct positive integers corresponding to coin values, also called denominations, in a real or imagined monetary system. For example, the coin system in common use in Canada is {1, 5, 10, 25, 100, 200}, where 1 corresponds to a 1-cent coin and 200 corresponds to a 200-cent (2-dollar) coin. For any coin system S, we assume that there is an unlimited supply of coins of each denomination, and we also assume that S contains 1,since this guarantees that any positive integer can be written as a sum of (possibly repeated) values in S.
Cashiers all over the world face (and solve) the following problem: For a given coin system and a positive integer amount owed to a customer, what is the smallest number of coins required to dispense exactly that amount? For example, suppose a cashier in Canada owes a customer 83 cents. One possible solution is 25+25+10+10+10+1+1+1, i.e.,8 coins, but this is not optimal, since the cashier could instead dispense 25 + 25 + 25 + 5 + 1 + 1 + 1, i.e., 7 coins (which is optimal in this case). Fortunately, the Canadian coin system has the nice property that the greedy algorithm always yields an optimal solution, as do the coin systems used in most countries. The greedy algorithm involves repeatedly choosing a coin of the
largest denomination that is less than or equal to the amount still owed, until the amount owed reaches zero. A coin system for which the greedy algorithm is always optimal is called canonical.
Your challenge is this: Given a coin system S = {c1, c2, . . . , cn }, determine whether S is canonical or non-canonical. Note that if S is non-canonical then there exists at least one counterexample, i.e., a positive integer x such that the minimum number of coins required to dispense exactly x is less than the number of coins used by the greedy algorithm. An example of a non-canonical coin system is {1, 3, 4}, for which 6 is a counterexample, since the greedy algorithm yields 4 + 1 + 1 (3 coins), but an optimal solution is 3 + 3 (2 coins). A useful fact (due to Dexter Kozen and Shmuel Zaks) is that if S is non-canonical, then the smallest counterexample is less than the sum of the two largest denominations.

输入

Input consists of a single case. The ﬁrst line contains an integer n (2 ≤ n ≤ 100), the number of denominations in the coin system. The next line contains the n denominations as space-separated integers c1 c2 . . . cn, where c1 = 1 and c1 < c2 < . . . < cn ≤ 106.

输出

Output “canonical” if the coin system is canonical, or “non-canonical” if the coin system is non-canonical.

样例输入

4
1 2 4 8

样例输出

canonical

题解：

给你一些货币的价值，假设给出一个价值，让你用已有的货币价值去组合成给出的价值，有两种策略，一种是每次都选最大的，最大的选不了再选次大的，直到凑齐；另一种是选择出组合出给出价值所需的最小货币数，问，是否存在这样一个价值，能让这两种策略选出的货币数量不相等，如果存在就是不规范的，不存在就是规范的。题目有一个提示，最小的反例小于两个最大面额的总和。所以我们就从1枚举到最大面额的两倍，先暴力求出第一种策略每个价值所需的数量，然后再用背包求出每个价值的最小货币表示数。然后一一比对，存在一个不相等就是不规范的了，训练赛的时候，一直WA，比赛结束的才注意到我们数组开小了Orz，希望以后不要在这么粗心啦。

AC代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<queue>
#pragma warning(disable:4996)
using namespace std;
typedef long long ll;
const int maxn = 2000005;
const int INF = 0x3f3f3f3f;
int n, m;
int a[maxn];
int dp[maxn];
int num[maxn];
int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &a[i]);}memset(dp, INF, sizeof(dp));dp[0] = 0;int sum = maxn;for (int i = 1; i <= maxn; i++) {int temp = i;int pos = n - 1;while(temp){if(temp >= a[pos]){num[i] += temp/a[pos];temp %= a[pos];}pos--;} }for (int i = 0; i < maxn; i++) {for (int j = 0; j < n; j++) {if (i - a[j] >= 0) {dp[i] = min(dp[i], dp[i - a[j]] + 1);}}}bool flag = false;for (int i = 1;i<a[n-1] * 2; i++) {if (dp[i] != num[i]) {printf("non-canonical\n");flag = true;break;}}if (!flag) {printf("canonical\n");}
}