Hdu 2546 meal card (Backpack)

 

Set the meal card balance to total
After analysis, we can conclude that the consumption of some meals should be as close as possible to total-5 yuan, and then we can purchase another meal to reach the overdraft...

It can prove that the last purchased food is the most valuable.
Proof

Set a1 a2 a3... an-1 an to set the maximum price for each meal

Set
Sum = total-5
A1 + a2 + a3 +... an-2 + an-1 + an = M
A1 + a2 + a3 +... + an-2 + an-1 = x1 add an (by the limit of 5 yuan) at this time excess (an-1)-(sum-x1) = an-sum + a1 + a2 +... + an-2 + an-1 RMB 1
A1 + a2 + a3 +... + an-2 + an = x2 add an-1 (by the limit of 5 yuan) at this time excess (an)-(sum-x2) = (an-1) -sum + a1 + a2 +... + an-2 + an RMB 2

1-2 formula = 2 * (an)-(an-1)> 0
So the 1-level excess is more
So I finally chose the food with the largest price.

Note: If the total balance is less than 5 yuan, you can directly output the total, but you cannot buy things at this time.

Dynamic Planning
01 total number of backpacks with problems-5 w for each backpack's value and weight

 

 

# Include <stdio. h>

# Include <string. h>
# Include <stdlib. h>
# Define max (a, B) a> B? A: B
Int cmp (const void * a, const void * B)
{
Return * (int *) a-* (int *) B;
}
Int main ()
{
Int n, a [5000], I, bb [5000], m, j;
While (scanf ("% d", & n), n! = 0)
{
Memset (bb, 0, sizeof (bb ));
For (I = 0; I <n; I ++)
Scanf ("% d", & a [I]);
Scanf ("% d", & m );
Qsort (a, n, sizeof (a [0]), cmp );
If (m <5)
{
Printf ("% d \ n", m );
Continue;
}
Memset (bb, 0, sizeof (bb ));
M = M-5;
Bb [0] = 0;
For (I = 0; I <n-1; I ++)
{
For (j = m; j> = a [I]; j --)
{
Bb [j] = max (bb [j], bb [j-a [I] + a [I]);
}
}
Printf ("% d \ n", m + 5-bb [m]-a [n-1]);
}
Return 0;

}