POJ 1014 Dividing（多重背包+二进制优化）

http://poj.org/problem?id=1014

题意：
6个物品，每个物品都有其价值和数量，判断是否能价值平分。

思路：

多重背包。利用二进制来转化成0-1背包求解。

 #include<iostream>
 #include<string>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 
 const int maxn = ;
 int sum;
 int a[];
 int d[maxn];
 int w[maxn];
 
 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     int kase = ;
     while (scanf("%d", &a[]))
     {
         for (int i = ; i <= ; i++)
             scanf("%d", &a[i]);
         sum = ;
         for (int i = ; i <= ; i++)
             sum += i*a[i];
         if (sum == )  break;
 
         printf("Collection #%d:\n", ++kase);
 
         if (sum % )
         {
             printf("Can't be divided.\n\n");
             continue;
         }
         int count = ;
         for (int i = ; i <= ; i++)
         {
             int m = a[i];
             int k = ;
             while (k < m)
             {
                 w[count++] = k*i;
                 m -= k;
                 k *= ;
             }
             w[count++] = m*i;
         }
         sum = sum / ;
         memset(d, , sizeof(d));
         for (int i = ; i < count; i++)
         {
             for (int j = sum; j >= w[i]; j--)
                 d[j] = max(d[j], d[j - w[i]] + w[i]);
         }
         if (d[sum]==sum)
             printf("Can be divided.\n\n");
         else
             printf("Can't be divided.\n\n");
     }
 }