Program3_U

 我现在做的是第三专题编号为1021的试题，具体内容如下所示：

Problem U

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 28   Accepted Submission(s) : 11
[align=left]Problem Description[/align]
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. <br>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.<br>
 

[align=left]Input[/align]
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. <br><br>The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.<br>
 

[align=left]Output[/align]
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.''. <br><br>Output a blank line after each test case.<br>
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

简单题意：

价值分别为1,2,3,4,5,6的物品的个数分别为 a[1],a[2],````a[6]

问能不能分成两堆价值相等的

解题思路：

   多种背包的问题，将其化简为01背包或完全背包。

编写代码：

#include<stdio.h>

#include<string.h>

int a[7];

int f[120005];

int v,k;

void ZeroOnePack(int cost,int weight)

{

    for(int i=v;i>=cost;i--)

       if(f[i-cost]+weight>f[i]) f[i]=f[i-cost]+weight;

}

void CompletePack(int cost,int weight)

{

    for(int i=cost;i<=v;i++)

        if(f[i-cost]+weight>f[i]) f[i]=f[i-cost]+weight;

}

void MultiplePack(int cost ,int weight,int amount)

{

    if(cost*amount>=v) CompletePack(cost,weight);

    else

    {

        for(int k=1;k<amount;)

        {

            ZeroOnePack(k*cost,k*weight);

            amount-=k;

            k<<=1;

        }

        ZeroOnePack(amount*cost,amount*weight);

    }

}

int main()

{

    int tol;

    int iCase=0;

    while(1)

    {

        iCase++;

        tol=0;

        for(int i=1;i<7;i++)

        {

            scanf("%d",&a[i]);

            tol+=a[i]*i;

        }

        if(tol==0) break;

        if(tol%2==1)

        {

            printf("Collection #%d:\nCan't be divided.\n\n",iCase);

            continue;

        }

        else

        {

            v=tol/2;

            memset(f,0,sizeof(f));

            for(int i=1;i<7;i++)

              MultiplePack(i,i,a[i]);

            if(f[v]==v)

              printf("Collection #%d:\nCan be divided.\n\n",iCase);

            else printf("Collection #%d:\nCan't be divided.\n\n",iCase);

        }

    }

    return 0;

}题意：价值分别为1,2,3,4,5,6的物品的个数分别为 a[1],a[2],````a[6]

//问能不能分成两堆价值相等的