有趣的换零钱问题

        经历一番波折终于写出来了, 在博客园记录一下~~

        给你若干钱, 有1元, 5元, 10元, 25元, 50元这5种小钱可以换. 求共有多少种方法.

#include <iostream>
using namespace std;
int s(int a)
{
     switch (a)
     {
     case 1: return 1; break;
     case 2: return 5; break;
     case 3: return 10; break;
    case 4: return 25; break;
    case 5: return 50; break;
    }
}
int way(int j, int k = 5)
{
    int s(int a);
    if (j == 0)
        return 1;
    else if ((j < 0)||(k == 0))
        return 0;
    else
        return way(j, k - 1) + way(j - s(k), k);
}
int main()
{    int p;
    int way(int j, int k =5);
    cout << "请输入钱数:";
    cin >> p;
    cout << way(p, 5);
}

        要是想一一列出1元, 5元, 10元, 25元, 50元各换了多少张, 再写一个程序.

#include <iostream>
using namespace std;
//1,5,10,25,50
int main()
{
    int a, i, j, k, o, p,q = 0;
    cin >> a;
    for (i = 0; i <= (int)a/50; i++)
    {for(j = 0; j <= (int)(a - 50*i)/25; j++)
    {for(k = 0; k <= (int)(a-50*i-j*25)/10; k++)
    {for(o = 0; o <= (int)(a-50*i-j*25-k*10)/5; o++)
    {for(p = 0; p <= (int)(a-50*i-j*25-k*10-o*5); p++)
    {
     if(50*i+25*j+10*k+5*o+p == a)
     {
          q++;
      cout <<"50的换"<<i<<"张,25的换"<<j<<"张,10的换"<<k<<"张,5的换"<<o<<"张,1块的换"<<p<<"张."<<endl;
    }}}}}}
    cout<<"共有:"<<q<<"种方法."<<endl;
}