hdu-1059 Dividing---多重背包

题目链接：

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

题目大意：

就是有价值1、2、3、4、5、6的硬币各多少个，然后让你判断能否把他们分成价值相等的两部分。

思路：

多重背包的模板

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 200000 + 10;
int n, m, cases;
int dp[maxn];
int sum;

void ZeroOnePack(int cost, int value)
{
    for(int v = sum; v >= cost; v--)//01背包正序
        dp[v] = max(dp[v], dp[v - cost] + value);
}

void CompletePack(int cost, int value)
{
    for(int v = cost; v <= sum; v++)
        dp[v] = max(dp[v], dp[v - cost] + value);
}
void MutiplePack(int cost, int value, int amount)
{
    if(cost * amount >= sum)//超过总体积，进行完全背包
        CompletePack(cost, value);
    else
    {
        int k = 1;//二进制优化
        while(k <= amount)
        {
            ZeroOnePack(k * cost, k * value);
            amount -= k;
            k *= 2;
        }
        ZeroOnePack(amount * cost, amount * value);
    }
}
int main()
{
    int a[10];
    while(cin >> a[1] >> a[2] >> a[3] >> a[4] >> a[5] >> a[6])
    {
        //if(a[1] + a[2] + a[3] + a[4] + a[5] + a[6] == 0)break;
        if(!a[1]&&!a[2]&&!a[3]&&!a[4]&&!a[5]&&!a[6])break;
        sum = 0;
        for(int i = 1; i <= 6; i++)sum += i * a[i];
        printf("Collection #%d:
",++cases);
        if(sum & 1)//总和为奇数，一定不能均分
        {
            printf("Can't be divided.

");
            continue;
        }
        memset(dp, 0, sizeof(dp));
        sum /= 2;
        for(int i = 1; i <= 6; i++)
            MutiplePack(i, i, a[i]);//多重背包
        if(dp[sum] == sum)
            printf("Can be divided.

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

");
    }
    return 0;
}