給你若干錢, 有1元, 5元, 10元, 25元, 50元這5種小錢可以換. 求共有多少種方法.
#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 << "請輸入錢數:"; |
cin >> p; |
cout << way(p, 5); |
} |
要是想一一列出1元, 5元, 10元, 25元, 50元各換了多少張, 再寫一個程序.
#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的換"<<i<<"張,25的換"<<j<<"張,10的換"<<k<<"張,5的換"<<o<<"張,1塊的換"<<p<<"張."<<endl; |
}}}}}} |
cout<<"共有:"<<q<<"種方法."<<endl; |
} |