Spoj-COINS-记忆化dp

Posted 2020-10-30 zzq

COINS - Bytelandian gold coins

#dynamic-programming

 

In Byteland they have a very strange monetary system.

Each Bytelandian gold coin has an integer number written on it. A coin n can be exchanged in a bank into three coins: n/2, n/3 and n/4. But these numbers are all rounded down (the banks have to make a profit).

You can also sell Bytelandian coins for American dollars. The exchange rate is 1:1. But you can not buy Bytelandian coins.

You have one gold coin. What is the maximum amount of American dollars you can get for it?

Input

The input will contain several test cases (not more than 10). Each testcase is a single line with a number n, 0 <= n <= 1 000 000 000. It is the number written on your coin.

Output

For each test case output a single line, containing the maximum amount of American dollars you can make.

Example

Input:
12
2

Output:
13
2

You can change 12 into 6, 4 and 3, and then change these into $6+$4+$3 = $13. If you try changing the coin 2 into 3 smaller coins, you will get 1, 0 and 0, and later you can get no more than $1 out of them. It is better just to change the 2 coin directly into $2.

 

      一开始开的1e9数组果断CE，然后开成1e8，对大于1e8的元素进行递归处理，小于1e8的将结果存到数组。

  

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long 
 4 const int maxn=100000000+10;
 5 LL f[maxn];
 6 LL g(LL n){
 7     if(n<=100000000) return f[n];
 8     return max(n,g(n/2)+g(n/3)+g(n/4));
 9 }
10 int main()
11 {
12     f[0]=0;
13     for(LL i=1;i<=maxn-10;++i){
14         f[i]=max(i,f[i/2]+f[i/3]+f[i/4]);
15     }
16     LL n;
17     while(scanf("%lld",&n)==1){
18         if(n<=100000000)
19         printf("%lld\n",f[n]);
20         else
21         printf("%lld\n",g(n));
22 
23     }
24     return 0;
25 }