hdu 1059 (多重背包) Dividing

这里;http://acm.hdu.edu.cn/showproblem.php?pid=1059

题意是有价值分别为1,2,3,4,5,6的商品各若干个,给出每种商品的数量,问是否能够分成价值相等的两份.

联想到多重背包,稍微用二进制优化一下。(最近身体不适,压力山大啊)

#include<iostream>
#include<cstring>
#include<cstdio>
#define inf 70000
using namespace std;
int dp[inf];
int sum;
void pack(int price)
{
    for(int i = sum; i >= price; i--) dp[i] = max(dp[i], dp[i - price] + price);
}
int main()
{
    int a[8],i,q=1,j;
    while (~scanf("%d %d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]))
    {
        if (a[1]==0&&a[2]==0&&a[3]==0&&a[4]==0&&a[5]==0&&a[6]==0)
            break;
        memset(dp,-inf,sizeof(dp));
        dp[0]=0;
        printf("Collection #%d:
",q++);
        sum=0;
        for (i=1;i<=6;i++)
            sum+=a[i]*i;
        if (sum%2!=0)
        {
            printf("Can't be divided.

");
            continue;
        }
        sum/=2;
        for (i=1;i<=6;i++)
        {
            if (i*a[i]>=sum)
            {
                for (j=i;j<=sum;j++)
                    dp[j]=max(dp[j],dp[j-i]+i);
            }
            else
            {
                int k = 1;
                while(k < a[i])
                {
                   pack(k * i);
                   a[i] -= k;
                   k += k;
                }
                pack(a[i] * i);
            }
        }
        if (dp[sum]==sum)
            printf("Can be divided.

");
        else
            printf("Can't be divided.

");
    }
    return 0;
}