(codechef) Bytelandian gold coins
2017-05-18 20:38
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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.
どこでもドア:https://www.codechef.com/problems/COINS
简单的DP,n<=11 时候 输出 n
n>11 dp
= dp[n/2]+dp[n/3]+dp[n/4]
code
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.
どこでもドア:https://www.codechef.com/problems/COINS
简单的DP,n<=11 时候 输出 n
n>11 dp
= dp[n/2]+dp[n/3]+dp[n/4]
code
#include<bits/stdc++.h> using namespace std; typedef long long LL; map<int , LL> dp; LL change(LL a){ if(a<=11) return a; if(dp.count(a)) return dp[a]; dp[a]=change(a/2)+change(a/3)+change(a/4); return dp[a]; } int main() { LL a; while(scanf("%lld",&a)!=EOF){ LL ans = change(a); printf("%lld\n",ans); } return 0; }
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