OJ Board

No. The problem is not like that.

For first example:

start:1 2 3 4 5 6 7 8 9 10

1st card:

1 2 . 4 5 6 7 8 9 10

1 2 . 4 5 . 7 8 9 10

1 2 . 4 5 . 7 8 . 10

2nd card:

1 2 . 4 5 . . 8 . 10

3rd card:

1 2 . 4 . . . 8 . 10

4th card:

1 2 . . . . . 8 . 10

5th card:

1 . . . . . . 8 . 10

1 ...

Because you have to check the result can be divide by people.

For example, 25.

(25-1)/3=8

so after first person, still 16 coconuts left.

(16-1)/3=5

so after second person, still 10 coconuts left.

and after third person, 6 coconuts left. ----> result

AND,

6 can be divide by 3, so 3 people is ...

The table at last is like this:
Table[0]=0
Table[1]=0
Table[2]=1
Table[3]=1
Table[4]=2
Table[5]=1
Table[6]=2
Table[7]=1
Table[8]=3
...
This means the number's "factors" --> I don't know whether this word or not

Example
8 = 2 * 2 * 2
So Table[8]=3

to FAQ:

Table[4]+Table[3]+Table[6]+Table[2 ...

Can anyone say where is wrong?
I try to solve it but fault :(
#define MAX 78599
int a[1000009];
int prime[MAX]={2,3,5},pows[168];
void calprime()
{
int i,j,k,flag;
for(i=3,j=7;i<MAX;j++,flag=1)
{
for(k=0;prime[k]*prime[k]<=j;k++)
if(j%prime[k]==0)
{
flag=0;
break;
}
if(flag)
prime[i ...

For example,
Your program print more.

Actually the non-self-numbers generated by this formula
are about 20 000 only. Which is only about 2% of all
numbers generated by that formula. But still their existence
makes the problem harder.

Thank you for your hint!
I got AC. :)
But it used about 6.7sec...
Is there a faster method easy ...

I already WA many times
So I want to test where my code go wrong
Can anyone post test cases for me?