poj 1014 Dividing（多重背包/母函数）

典型的均分问题，最简单的方法是用多重背包，当然也可以用母函数解决，但是这题n很大，最大是20000，所以用母函数必须剪枝。

但是不会剪枝，看到有好多剪枝方法...最多的一个是将数量6作为分界线。。虽然也没怎么看懂。。。贴地址：剪枝方法

多重背包代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e5+5;
int v[maxn], d[maxn], V[maxn], dp[maxn];
int main(void)
{
    int ca = 1;
    for(int i = 1; i <= 6; i++) v[i] = i;
    while(1)
    {
        int sum = 0;
        for(int i = 1; i <= 6; i++)
            scanf("%d", &d[i]), sum += d[i]*v[i];
        if(!sum) break;
        printf("Collection #%d:\n", ca++);
        if(sum%2) puts("Can't be divided.\n");
        else
        {
            sum /= 2;
            int k = 1;
            memset(dp, 0, sizeof(dp));
            for(int i = 1; i <= 6; i++)
            {
                for(int j = 1; j <= d[i]; j*=2)
                {
                    V[k++] = j*v[i];
                    d[i] -= j;
                }
                if(d[i] > 0) V[k++] = d[i]*v[i];
            }
            for(int i = 0; i < k; i++)
                for(int j = sum; j >= V[i]; j--)
                    dp[j] = max(dp[j], dp[j-V[i]]+V[i]);
            if(dp[sum] == sum) puts("Can be divided.\n");
            else puts("Can't be divided.\n");
        }
    }
    return 0;
}

母函数代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 2*1e5+5;
int a[10], c1[maxn], c2[maxn];
int main(void)
{
    int ca = 1;
    while(1)
    {
        int sum = 0;
        for(int i = 1; i <= 6; i++)
        {
            scanf("%d", &a[i]);
            if(a[i] > 6)
            {
                if(a[i]%2) a[i] = 5;
                else a[i] = 6;
            }
            sum += a[i]*i;
        }

        if(!sum) return 0;
        printf("Collection #%d:\n", ca++);
        if(sum%2) printf("Can't be divided.\n");
        else
        {
            memset(c1, 0, sizeof(c1));
            memset(c2, 0, sizeof(c2));
            for(int i = 0; i <= a[1]; i++) c1[i] = 1;
            for(int i = 2; i <= 6; i++)
            {
                if(!a[i]) continue;
                for(int j = 0; j <= sum/2; j++)
                    for(int k = 0; j+k <= sum/2 && k/i <= a[i]; k += i)
                        c2[j+k] += c1[j];
                for(int j = 0; j <= sum/2; j++)
                    c1[j] = c2[j], c2[j] = 0;
            }
            printf("%s\n", c1[sum/2] ? "Can be divided." : "Can't be divided.");
        }
        printf("\n");
    }
    return 0;
}

Dividing

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 69280		Accepted: 18068

Description

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

Input

Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , n6 , where ni is the number of marbles of value i. So, the example from above would be described by the input-line "1 0 1 2 0 0". The maximum total number of marbles will be 20000. 
The last line of the input file will be "0 0 0 0 0 0"; do not process this line.

Output

For each collection, output "Collection #k:", where k is the number of the test case, and then either "Can be divided." or "Can't be divided.". 
Output a blank line after each test case.

Sample Input

1 0 1 2 0 0 
1 0 0 0 1 1 
0 0 0 0 0 0 

Sample Output

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

Source

Mid-Central European Regional Contest 1999