我的创作纪念日 CF1620D Exact Change 题解

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突然发现自己在 CSDN 已经创作了 $128$ 天，时间流逝得好快。

在这个特殊的日子里，发篇题解纪念一下。

link

题面翻译

• 有 $n$ 个物品，第 $i$ 个物品价格为 $a_i$。
• 有三种货币，面值分别为 $1,2,3$。
• 求最小需要多少张货币，才能表出所有的 $a_i$。
• 多组测试数据。
• $1\le t\le 1000,1\le n\le 100,1\le a_i\le 10^9$。

题目描述

One day, early in the morning, you decided to buy yourself a bag of chips in the nearby store. The store has chips of $n$ different flavors. A bag of the $i$ -th flavor costs $a_i$ burles.

The store may run out of some flavors, so you’ll decide which one to buy after arriving there. But there are two major flaws in this plan:

1. you have only coins of $1$ , $2$ and $3$ burles;
2. since it’s morning, the store will ask you to pay in exact change, i. e. if you choose the $i$ -th flavor, you’ll have to pay exactly $a_i$ burles.

Coins are heavy, so you’d like to take the least possible number of coins in total. That’s why you are wondering: what is the minimum total number of coins you should take with you, so you can buy a bag of chips of any flavor in exact change?

输入格式

The first line contains a single integer $t$ ( $1\le t\le 1000$ ) — the number of test cases.

The first line of each test case contains the single integer $n$ ( $1\le n\le 100$ ) — the number of flavors in the store.

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n$ ( $1\le a_i\le 10^9$ ) — the cost of one bag of each flavor.

输出格式

For each test case, print one integer — the minimum number of coins you need to buy one bag of any flavor you’ll choose in exact change.

样例 #1

样例输入 #1

4
1
1337
3
10 8 10
5
1 2 3 4 5
3
7 77 777

样例输出 #1

446
4
3
260

提示

In the first test case, you should, for example, take with you $ 445 $ coins of value $3$ and $1$ coin of value $2$ . So, $1337=445\cdot 3+1\cdot 2$ .

In the second test case, you should, for example, take $ 2 $ coins of value $3$ and $2$ coins of value $2$ . So you can pay either exactly $8=2\cdot 3+1\cdot 2$ or $10=2\cdot 3+2\cdot 2$ .

In the third test case, it’s enough to take $1$ coin of value $3$ and $2$ coins of value $1$ .

思路

注意到货币的面值只能是 $1,2,3$，就从这一点特殊性质入手。

可以发现，面值为 $1$ 的纸币不会超过 $2$ 张，因为 $2$ 张 $1$ 元纸币相当于 $1$ 张 $2$ 元纸币，显然后者更优。

同理，面值为 $2$ 的纸币不会超过 $3$ 张，因为 $3$ 张 $2$ 元纸币相当于 $2$ 张 $3$ 元纸币，显然后者更优。

然后暴力枚举即可~

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll N=2e5+10,inf=1e9;
ll T,n,a[N],res;
ll solve(ll t,ll t1,ll t2){ll ct=inf; for(int i=0;i<=t1;i++)//面值为 1 的纸币数量for(int j=0;j<=t2;j++){//面值为 2 的纸币数量ll tmp=t-i-j*2;//剩余钱数if(tmp<0) continue;if(!(tmp%3)) ct=min(ct,tmp/3);//面值为 3 的纸币数量}return ct;
}
ll work(ll t1,ll t2){ll ans=-1;//面值为 3 的纸币数量最大值for(int i=1;i<=n;i++) ans=max(ans,solve(a[i],t1,t2));return ans+t1+t2;
}
int main(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cin>>T;while(T--){cin>>n;res=inf;for(int i=1;i<=n;i++) cin>>a[i];for(int i=0;i<=1;i++)for(int j=0;j<=2;j++) res=min(res,work(i,j));cout<<res<<"\n";}return 0;
}

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