For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.
Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.
#include<iostream>
using namespace std;
const int money=251;
const int coin=101;
int dp[money][coin]={0}; //dp[i][j]表示金额为i，硬币数为j的种类方法
int value[5]={1,5,10,25,50};
void solve()
{
//for(int i=0;i<money;i++) 这里为什么只能dp[0][0]=0,是因为dp[j][1]=dp[j][1]+dp[j-value[i]][k-1],只有j-value[i]==0时，才能在加1
//dp[i][0]=1;
dp[0][0]=1;
for(int i=0;i<5;i++)
{
for(int j=value[i];j<money;j++)
{
for(int k=1;k<coin;k++)
{
dp[j][k]=dp[j][k]+dp[j-value[i]][k-1];
}
}
}
}
int main()
{
solve();
int ans[money]={0};
ans[0]=1; //注意这一步，一定不能忘，一个特殊值
for(int i=1;i<money;i++)
{
for(int j=1;j<coin;j++)
{
ans[i]=dp[i][j]+ans[i];
}
}
int s;
while(cin>>s)
cout<<ans[s]<<endl;
return 0;
}
因为刚开始学习动态规划，这种硬币问题算是入门的问题。到现在为止，我所理解的动态规划是利用递推关系式，对于某个解值是需要用之前的算出来的进行求解，也就是类似于我要想知道第三个值的数，就需要让第一个数乘以第二个数，举个例子，就像我们都熟知的斐波那契数列，但要注意初始值一定要弄明白，否则递推关系式将进行不下去。另外斐波那契数列是最简单的dp,稍微复杂一点的需要进行不止一次地递推，就像这次的硬币问题，需要递推5次。