A. Secrets

Time Limit: 20 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/333/problem/A

Description

Gerald has been selling state secrets at leisure. All the secrets cost the same: n marks. The state which secrets Gerald is selling, has no paper money, only coins. But there are coins of all positive integer denominations that are powers of three: 1 mark, 3 marks, 9 marks, 27 marks and so on. There are no coins of other denominations. Of course, Gerald likes it when he gets money without the change. And all buyers respect him and try to give the desired sum without change, if possible. But this does not always happen.

One day an unlucky buyer came. He did not have the desired sum without change. Then he took out all his coins and tried to give Gerald a larger than necessary sum with as few coins as possible. What is the maximum number of coins he could get?

The formal explanation of the previous paragraph: we consider all the possible combinations of coins for which the buyer can not give Gerald the sum of n marks without change. For each such combination calculate the minimum number of coins that can bring the buyer at least n marks. Among all combinations choose the maximum of the minimum number of coins. This is the number we want.

Input

The single line contains a single integer n (1 ≤ n ≤ 10¹⁷).

Please, do not use the %lld specifier to read or write 64 bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

In a single line print an integer: the maximum number of coins the unlucky buyer could have paid with.

Sample Input

1

Sample Output

1

HINT

题意

所有的硬币面额都是3的幂,给出价格N,假设N无法由目前所有的硬币组合成,那么求在多花最少钱的情况下最多用多少枚硬币.

题解：

N不断除3，直到N不是3的倍数，答案就是n/3+1

代码：

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                   typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 200001#define mod 10007#define eps 1e-9int Num;char CH[20];//const int inf=0x7fffffff; //нчоч╢Сconst int inf=0x3f3f3f3f;/*inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}*/inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}//**************************************************************************************ll n;int main(){ n=read(); while(n%3==0) n/=3; cout<

转载于:https://www.cnblogs.com/qscqesze/p/4528529.html

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