HDU1059 Dividing(多重背包，二进制优化，模板题)

题目：

Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24745    Accepted Submission(s): 7052

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 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.

思路：

题意是很多石头，按照等级划分为1--6，给的数据是第几个等级有几个石头，问的是最后能不能平分，不好意思，用了模板

代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int dp[50000],v;void bag01(int c,int w)//01背包{    int i;    for(i=v; i>=c; i--)    {        if(dp[i]<dp[i-c]+w)        {            dp[i]=dp[i-c]+w;        }    }}void bagall(int c,int w)//完全背包{    int i;    for(i=c; i<=v; i++)    {        if(dp[i]<dp[i-c]+w)        {            dp[i]=dp[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 a[10],q=1;    while(1)    {        mem(dp,0);        int flag=0,sum=0;        for(int i=0; i<6; i++)        {            scanf("%d",&a[i]);            if(a[i]==0)flag++;            else sum+=a[i]*(i+1);        }        if(flag==6)return 0;        if(sum&1)//如果是奇数，直接输不能            printf("Collection #%d:\nCan't be divided.\n\n",q++);        else        {            v=sum/2;            for(int i=0; i<6; i++)                if(a[i])                    multbag(i+1,i+1,a[i]);//进行多重背包            if(v==dp[v])                printf("Collection #%d:\nCan be divided.\n\n",q++);            else                printf("Collection #%d:\nCan't be divided.\n\n",q++);        }    }    return 0;}