poj 1014 Dividing（多重背包）

题意：有很多的 marble 每种 marble有一个val 1 ...6 。问给否把它们分成两份，使得两份的val相同。

思路：多重背包 算出总的val的一半，如果dp[val]==val/2 那就能dividing

//1844K   0MS

#include <stdio.h>
#include <string.h>
#define Max 430000
const int n = 6;
int dp[Max];
int count[n],val;
int max (int a,int b)
{
    return a> b ? a : b;
}

void MulPack(int cost,intamount)   //多重背包
{
    int i;
    if(cost*amount >= val)
    {
       for (i = cost; i <= val; i ++)
           dp[i] = max (dp[i],dp[i-cost] + cost);
       return ;
    }
    int k =1;
    while (k< amount)
    {
       for (i = val; i >= k*cost; i --)
           dp[i] = max (dp[i],dp[i - k*cost] + k*cost);
       amount -= k;
       k *= 2;
    }
    for (i =val; i >= amount*cost; i --)
       dp[i] = max (dp[i],dp[i - amount*cost] + amount*cost);
}
int main ()
{
    int i,sum,t= 0;;
    while(1)
    {
       sum = 0,val = 0;
       memset (dp,0,sizeof(dp));
       for (i = 0; i < 6; i ++)
       {
           scanf ("%d",&count[i]);
           val += count[i]*(i+1);
           if (!count[i])
               sum ++;
       }
       if (sum == n)
               break;
       printf ("Collection #%d:\n",++t);
       if (val%2 ==1)            //值为奇的话，就直接不可能
           printf ("Can't be divided.\n");
       else
       {
           val /= 2;
           for (i = 0; i < n; i ++)
               MulPack (i+1,count[i]);

           if (dp[val] == val)
               printf ("Can be divided.\n");
           else
               printf ("Can't be divided.\n");
       }
       printf ("\n");
    }
    return0;
}