Dividing

                                      Dividing

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 be ``0 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 either ``Can be divided.'' or ``Can't be divided.''. 

Output a blank line after each test case. 

 

Sample Input

 1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0 

 

Sample Output

 Collection #1:Can't be divided.Collection #2:Can be divided.
题目的意思是有价值为一到六的石头，输入的数为对应的石头的个数，让你将这些石头平均分成相同价值的两份（石头不能分割），如果无法办到就输出“Can't be divided.”，如果能分就输出“Can be divided.”。
这道题我是直接用模板做的。。。。
代码：
#include<stdio.h>#include<iostream>#include<string.h>using namespace std;int d[120005],a[7],v;void bag01(int c,int w)//01背包{    int i;    for(i=v; i>=c; i--)    {        if(d[i]<d[i-c]+w)        {            d[i]=d[i-c]+w;        }    }}void bagall(int c,int w)//完全背包{    int i;    for(i=c; i<=v; i++)    {        if(d[i]<d[i-c]+w)        {            d[i]=d[i-c]+w;        }    }}void multbag(int c,int w,int n)//多重背包{    if(c*n>=v)    {        bagall(c,w);        return ;    }    int k=1;    while(k<=n)    {        bag01(k*c,k*w);        n=n-k;        k=k*2;    }    bag01(n*c,n*w);}int main(){    int t=1;    while(cin>>a[1]>>a[2]>>a[3]>>a[4]>>a[5]>>a[6])  //输入对应石头的个数    {        int sum=a[1]*1+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6;  //算出总价值        if(sum==0)break;        printf("Collection #%d:\n",t++);           if(sum%2==1)    //如果总价值为奇数，无法平均分成两份，直接输出Can't....        {            printf("Can't be divided.\n\n");            continue;        }        v=sum/2;        memset(d,0,sizeof(d));        for(int i=1;i<7;i++)            if(a[i])    //如果有价值为i的石头            multbag(i,i,a[i]);       if(d[v]==v)    //如果能平均分成两份        printf("Can be divided.\n\n");       else        printf("Can't be divided.\n\n");    }    return 0;}