Poj 1742 Coins (Backpack)

The number of coins with n denominations and the number of each of these coins must be calculated (less than or equal to m)

Ps: the method of using multiple backpacks timed out on poj (line hdoj), and then I saw a method on the Internet, use f [j] to indicate the sum of money that can constitute the value of j (0 and 1 ), use used [j] to indicate how many times A [I] is used when the current face value is j.
Code:
# Include <stdio. h>
# Include <string. h>
Int main ()
{
Int I = 0, j = 0, n = 0, m = 0, ans = 0, A [102], C [102], f [100002], used [100002];
While (scanf ("% d", & n, & m), n + m)
{
Memset (f, 0, sizeof (f ));
For (I = 0; I <n; I ++)
Scanf ("% d", & A [I]);
For (I = 0; I <n; I ++)
Scanf ("% d", & C [I]);
F [0] = 1;
For (I = 0; I <n; I ++)
{
For (j = 0; j <= m; j ++) // all initialization, indicating that A [I] 0 times is used when the current face value is j.
Used [j] = 0;
For (j = A [I]; j <= m; j ++)
{
If (! F [j] & f [j-A [I] & used [j-A [I] <C [I]) // when f [j] is 1, used [j] = 0
 
{
F [j] = 1;
Used [j] = used [j-A [I] + 1;
}
}
}
Ans = 0;
For (I = 1; I <= m; I ++)
If (f [I])
Ans ++;
Printf ("% d \ n", ans );
}
Return 0;
}

Multiple backpacks
Code:
# Include <stdio. h>
# Include <string. h>
Int n = 0, m = 0, A [102], C [102], dp [100002];
Void zero_one (int weight, int value)
{
Int j = 0;
For (j = m; j> = weight; j --)
Dp [j] = dp [j]> dp [j-weight] + value? Dp [j]: dp [j-weight] + value;
}
Int main ()
{
Int I = 0, j = 0, k = 0, sum = 0, ans = 0;
While (scanf ("% d", & n, & m), n + m)
{
Memset (dp, 0, sizeof (dp ));
Ans = 0;
For (I = 0; I <n; I ++)
Scanf ("% d", & A [I]);
For (I = 0; I <n; I ++)
Scanf ("% d", & C [I]);
For (I = 0; I <n; I ++)
{
Sum = A [I] * C [I];
If (sum> m) // full backpack
For (j = A [I]; j <= m; j ++)
Dp [j] = dp [j]> dp [j-A [I] + A [I]? Dp [j]: dp [j-A [I] + A [I];
Else
{
K = 1;
While (k <C [I]) // binary splitting
{
Zero_one (k * A [I], k * A [I]);
C [I]-= k;
K * = 2;
}
Zero_one (C [I] * A [I], C [I] * A [I]);
}
}
For (I = 1; I <= m; I ++)
If (dp [I] = I)
Ans ++;
Printf ("% d \ n", ans );
}
Return 0; www.2cto.com
}

Author: ulquiorra0cifer