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[],i,q=,j;
     while (~scanf("%d %d %d %d %d %d",&a[],&a[],&a[],&a[],&a[],&a[]))
     {
         if (a[]==&&a[]==&&a[]==&&a[]==&&a[]==&&a[]==)
             break;
         memset(dp,-inf,sizeof(dp));
         dp[]=;
         printf("Collection #%d:\n",q++);
         sum=;
         for (i=;i<=;i++)
             sum+=a[i]*i;
         if (sum%!=)
         {
             printf("Can't be divided.\n\n");
             continue;
         }
         sum/=;
         for (i=;i<=;i++)
         {
             if (i*a[i]>=sum)
             {
                 for (j=i;j<=sum;j++)
                     dp[j]=max(dp[j],dp[j-i]+i);
             }
             else
             {
                 int k = ;
                 while(k < a[i])
                 {
                    pack(k * i);
                    a[i] -= k;
                    k += k;
                 }
                 pack(a[i] * i);
             }
         }
         if (dp[sum]==sum)
             printf("Can be divided.\n\n");
         else
             printf("Can't be divided.\n\n");
     }
     return ;
 }