I think that's indeed the correct answer.

For any 4-point group we have 4 combinations, and for any 5-point groups we have 5. So there are
4 * C(7 ,4) + 5 * C(7, 5) = 245.

For any 6-point groups, as N = 7, we could know that it is a complete group for all 7 points with only one missing. So the 7 "different" groups are actually the same.

Let's count how many interesting triangles in one group. For they are all the same, we take point 1, 2, 3, 4, 5, 6 on the circle, which is in CCW order (7 is missing).

We have the following two ways to connect them and make them intersect pairwise:
(1-4, 2-5, 3-6): a small triangle is formed inside the circle.
and
(1-3, 2-5, 4-6): a long triangle is formed, with 2 of its vertices inside the circle.

So, there are 2 (or, at least 2) combinations for every 6-point groups in N = 7. So the total number will be
245 + 7 * 2 = 259
(or, at least 259). But all the ACed program will give the answer 252.

Why? Where did I have a mistake? Anyone, please, tell me the truth about this problem.

Thanks again in advance!

3 of your ways to connect 6 points are wrong:
(1-4,2-6,3-5) - segments 2-6 and 3-5 don't intersect
(2-5,1-3,4-6) - segments 1-3 and 4-6 don't intersect
(3-6,1-5,2-4) - segments 1-5 and 2-4 don't intersect

It is true, because any six different points construct a CONVEX 6-gon if they lie on one circle.

Видимо, по своему желанию может ходить за исключением 1-го и n-ого постамента

i have WA #6 can u give me some test?

нет, это видно из примера

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>

char* add(char *p1, char *p2)
{
    char *p;
    char temp=0, end1=0, end2=0;
    int len,i;
    len=strlen(p1)>strlen(p2)?strlen(p1)+2:strlen(p2)+2;
    p=(char *)malloc(len*sizeof(char));
    for(i=0; i<len-1; ++i)
    {
        if(!p1[i]) end1=1;
        if(!p2[i]) end2=1;
        if(end1&&end2&&!temp) break;
        if(!end1) temp+=(p1[i]-'0');
        if(!end2) temp+=(p2[i]-'0');
        p[i]=temp%10+'0';
        temp/=10;
    }
    p[i]=0;
    while(p[--i]=='0'&& &p[i]!=p) p[i]=0;
    return p;
}

char* sub(char *p1, char *p2)
{
    char *p;
    char temp=0, end1=0, end2=0;
    int len,i;
    len=strlen(p1)>strlen(p2)?strlen(p1)+2:strlen(p2)+2;
    p=(char *)malloc(len*sizeof(char));
    for(i=0; i<len-1; ++i)
    {
        if(!p1[i]) end1=1;
        if(!p2[i]) end2=1;
        if(end1&&end2&&!temp) break;
        if(!end1) temp+=(p1[i]-'0');
        if(!end2) temp-=(p2[i]-'0');
        p[i]=(temp+100)%10+'0';
        if(temp<0)
        {
            temp/=10;
            temp--;
        }
        else temp=0;
    }
    p[i]=0;
    while(p[--i]=='0'&& &p[i]!=p) p[i]=0;
    return p;
}

char* mul(char *p1, int k)
{
    char *p;
    char temp=0, end1=0;
    int len,i;
    len=strlen(p1)+2;
    p=(char *)malloc(len*sizeof(char));
    if(k==10)
    {
        p[0]='0';
        p[1]=0;
        strcat(p,p1);
        return p;
    }
    for(i=0; i<len-1; ++i)
    {
        if(!p1[i]) end1=1;
        if(end1&&!temp) break;
        if(!end1) temp+=(p1[i]-'0')*k;
        p[i]=temp%10+'0';
        temp/=10;
    }
    p[i]=0;
    while(p[--i]=='0'&& &p[i]!=p) p[i]=0;
    return p;
}

void printb(char *p)
{
    int i;
    for(i=strlen(p)-1;i>=0; i--) printf("%c", p[i]);
}

int main()
{
    int N, K;
    int i;
    char* result, *t1, *t2;
    char* Nulls[2000];

    printf("Enter N and K: ");
    scanf("%d%d",&N,&K);

    Nulls[0]="0";
    result=(char *)malloc(2*sizeof(char));
    result[0]=K-1+'0';
    result[1]=0;

    try
    {

    for(i=1; i<N;++i)
    {
        Nulls[i]=sub(result,Nulls[i-1]);
        t1=mul(Nulls[i],K);
        t2=mul(Nulls[i-1],(K-1));
        result=add(t1,t2);
        free(t1);
        free(t2);
    }

    }
    catch(...)
    {
        printf("Error");
        while(!kbhit());
        return 0;
    }

    printf("Result = ");
    printb(result);
    free(result);
    while(!kbhit());
    return 0;
}

for N=5

2
1 2
1 3

Use long not int...Look at my code

Thank you! I got AC (0.125sec)