hdu 1059 Dividing

Dividing

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 43   Accepted Submission(s) : 19

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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.
1-6元钱 给出每个的数量 问能否凑成一半的价格求和若是奇数肯定不行然后再初始化dp数组全为0，令dp[0]=1，代表凑i元若是1表示可以，若是0则表示不可以，所以在循环中不是取最大了而是令能取到的i的dp[i]为1，最终就知道能否凑出所求的了，这里要注意，在赋值1的时候要先判断取这枚硬币之前的金钱量能否取到。
#include<iostream>#include <cstdio>  #include <algorithm>  #include <cstring>  using namespace std;  int main()  {      int a[11],sum,v,i,j,k,cnt,js = 1;      int dp[111111];    while(~scanf("%d",&a[1]))      {          sum = a[1];          for(i = 2;i<=6;i++)          {              cin>>a[i];             sum+=a[i]*i;         }          if(sum==0)          break;          cout<<"Collection #"<<js<<":"<<endl;        js++;        if(sum%2==1)        {          cout<<"Can't be divided."<<endl<<endl;             continue;          }          v = sum/2;          memset(dp,0,sizeof(dp));          dp[0] = 1;          for(i = 1;i<=6;i++)          {              if(a[i]==0)              continue;              for(j = 1;j<=a[i];j=j*2)            {                  cnt = j*i;                  for(k = v;k>=cnt;k--)                  {                      if(dp[k-cnt])                    dp[k] = 1;                  }                  a[i]-=j;              }              cnt = a[i]*i;            if(cnt)              {                  for(k = v;k>=cnt;k--)                  {                      if(dp[k-cnt])                      dp[k] = 1;                  }              }          }          if(dp[v])          cout<<"Can be divided."<<endl<<endl;         else          cout<<"Can't be divided."<<endl<<endl;     }        return 0;  }