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 = 70000;
int sum;
int a[7];
int d[maxn];
int w[maxn];

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int kase = 0;
    while (scanf("%d", &a[1]))
    {
        for (int i = 2; i <= 6; i++)
            scanf("%d", &a[i]);
        sum = 0;
        for (int i = 1; i <= 6; i++)
            sum += i*a[i];
        if (sum == 0)  break;

        printf("Collection #%d:
", ++kase);

        if (sum % 2)
        {
            printf("Can't be divided.

");
            continue;
        }
        int count = 0;
        for (int i = 1; i <= 6; i++)
        {
            int m = a[i];
            int k = 1;
            while (k < m)
            {
                w[count++] = k*i;
                m -= k;
                k *= 2;
            }
            w[count++] = m*i;
        }
        sum = sum / 2;
        memset(d, 0, sizeof(d));
        for (int i = 0; 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.

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

");
    }
}