I sent your formula and got AC. Try to output answer so:

cout << fixed << setprecision(0) << z;

or so:

printf("%.0lf",z);

And you must use double instead of float

try this test

jkhdSaFKrojkllSafkroppp

answer 10

at first my answer is 20.I saw my bug with this test.
you must find max the same letter with "Sandro" and max lowercase letter.

Edited by author 18.10.2010 18:04

Edited by author 18.10.2010 18:04

Edited by author 18.10.2010 18:07

Edited by author 18.10.2010 18:08

It was possible only in 2008 and earlier.

Sorry, for this example, the correct output is T = 4 !!!
Check your algorithm again !!!
mailto : trungduck@yahoo.com

> For example:
> 7 10
> 1 2
> 2 3
> 2 4
> 2 5
> 3 5
> 4 5
> 5 6
> 5 7
> 6 7
> 1 7
> At first the DFS procedure find the loop:
> (1,2,3,5,6,7,1),then it'll find (1,2,3,5,7,1).
> and next,no loop with unused edge will be found!?!
> In fact,the maximum T is 3(it's obvious),but use greedy
> method + DFS the number of T only reach 2.
> Unless we regulated the order of search
> (it means DFS can't be right!),otherwise we have to
> delete a found loop to let more loop to have unused
> edge,then how to make the regulations?
>
>

I guess the 10-th test checks for your attintive reading of the condition: "1. The first LETTER in a sentence is small."

For example, the text

"Hello! 238 my Friend"

has an error, because the first letter in the second sentence is NOT the digit '2' but the letter 'm' which must be capital.

Because first chislo is m (don't n)

Edited by author 17.10.2010 00:31

Edited by author 17.10.2010 00:31

Problems are added.

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;
}