The currency system of country Z includes 4 kinds of coins with a face value of 1, 4, 16 and 64 yuan, as well as a banknote with a face value of 1024 yuan. Now Xiao Y uses 1024 yuan banknotes to buy a product with a value of N (0 <N \le 1024)N (0<N≤1024), how many coins will he receive at least?
Enter a description:
一行,包含一个数N。
Output description:
一行,包含一个数,表示最少收到的硬币数。
Example:
输入
200
输出
17
说明
花200,需要找零824块,找12个64元硬币,3个16元硬币,2个4元硬币即可。
Remarks:
对于100%的数据,N (0 < N \le 1024)N(0<N≤1024)。
Analysis: Greedy is not necessarily right, it is recommended to use dynamic programming
#include <iostream>
#include <vector>
using namespace std;
int min(int a,int b)
{
return (a < b) ? a : b;
}
int ans(int num)
{
//为了取最小,先塞最大的给数组,可计算0~num的找零
vector<int> dp(num + 1, num + 1);
//初始化0块钱
dp[0] = 0;
//建立货币系统
int money[] = {
1,4,16,64 };
//从1块钱开始,到第num块钱
for(int i=1;i<=num;i++)
{
//从塞1块钱开始,慢慢塞大钱以取最小值
for(int j=0;j<4;j++)
{
//看现在的钱可以用什么样的货币支付
if(i>=money[j])
{
dp[i] = min(dp[i], dp[i - money[j]] + 1);
}
}
}
return dp[num];
}
int main()
{
int N;//商品价格
cin >> N;
cout << ans(1024 - N) << endl;
return 0;
}