hdu1059 Dividing（生成函数||背包）

Problem 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 describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, …, 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 be0 0 0 0 0 0”; do not process this line.

 

Output

For each colletcion, output Collection #k:'', where k is the number of the test case, and then eitherCan 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.
    
    
    
    

Statistic |  Submit |  Discuss | Note

分析：
生成函数
但是我们没法直接化式子
然而我们可以像这道题一样暴力处理

我们运用背包的思维，倒序处理：

c[0]=1;
for (int i=1;i<=6;i++)
if (num[i])
{
    for (int j=s;j>=0;j--)
    if (c[j])
        for (int k=i;k<=num[i]*i&&k<=s;k+=i)
            c[j+k]|=c[j];
}

tip

我们把每一种marble的数量%10，就可以大大缩小数据范围（不知道是什么原理）

疯狂WA，都不知道怎么回事
后来我把一个小优化去掉了，就莫名其妙的A了

//这里写代码片
#include
#include
#include

using namespace std;

const int N=20010;
int num[10],c[N];

int main()
{
    int s=0,cnt=0;
    for (int i=1;i<=6;i++) {
        scanf("%d",&num[i]);
        num[i]%=10; s+=i*num[i];    //这个%10很神奇，缩小了数据量
    }

    while (s)
    {
        if (s&1)
        {
            printf("Collection #%d:\nCan't be divided.\n",++cnt);
            puts("");
            s=0;
            for (int i=1;i<=6;i++) {
                scanf("%d",&num[i]);
                num[i]%=10; s+=i*num[i];
            }
            continue;
        }

        memset(c,0,sizeof(c));

        c[0]=1;
        for (int i=1;i<=6;i++)
        if (num[i])
        {
            for (int j=s;j>=0;j--)
            if (c[j])
                for (int k=i;k<=num[i]*i&&k<=s;k+=i)
                    c[j+k]|=c[j];
        }

        if (c[s/2])
            printf("Collection #%d:\nCan be divided.\n",++cnt);
        else printf("Collection #%d:\nCan't be divided.\n",++cnt);
        puts("");

        s=0;
        for (int i=1;i<=6;i++) {
            scanf("%d",&num[i]);
            num[i]%=10; s+=i*num[i];
        }
    }
    return 0;
}