http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1005&ojid=0&cid=3900
Dividing
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 8 Accepted Submission(s) : 2
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.
The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.
Output a blank line after each test case.
#include<cstdio>
#include<cstring>
using namespace std;
int dp[60500];
int main()
{
int n[7];
int v[7];
int sum=0;
int test=1;
while(scanf("%d",&n[1])!=EOF)
{
memset(dp,0,sizeof(dp));
int sum=n[1];
int t;
for(int i=2; i<=6; i++)
{
scanf("%d",&t);
n[i]=t%10;
sum+=i*n[i];
}
if(n[1]==0&&n[2]==0&&n[3]==0&&n[4]==0&&n[5]==0&&n[6]==0) break;
if(sum%2==1)
{
printf("Collection #%d:\nCan't be divided.\n\n",test);
test++;
continue;
}
for(int i=1; i<=6; i++)
v[i]=i;
int V=sum/2;
dp[0]=1;
for(int i=1; i<=6; i++)
{
for(int j=0; j<n[i]; j++)
{
for(int k=V; k>=v[i]; k--)
{
if(dp[k-v[i]])
dp[k]=1;
}
}
}
// for(int i=V;i>=0;i--)
//printf("%d %d\n",dp[i],i);
if(dp[V]==1)
{
printf("Collection #%d:\nCan be divided.\n\n",test);
test++;
}
else
{
printf("Collection #%d:\nCan't be divided.\n\n",test);
test++;
}
}
return 0;
}