I got AC with O(n^3) solution, based on DP.
I suppose, that testcases are too weak. Anyway, I think, that it's not normal to get AC for O(n^3) solution.
)

My mail: bsc98001@yahoo.com

for N,

####...N/2times ####...N/2 times
consider first half N/2 digits

sum(digit root) will always be from 0 to 9
it will be 0 only if all digits are 0's
and it'll will 1,2,3 .... ,9 in other cases...

for N=4
first half will have 2 digits ( total 100 numbers )
1 number(00) will have digit root = 0
99 numbers will have digit root other than 0
( i.e 11 numberes --> digit root 1 , 11 numbers ---> digit root 2 etc )

so for N=4 ,
case i : first half = 00 second half =00 -----> n1 = 1
case ii: first half --> 99 ways , second half --> 11 ways  n2= 99*11

total number = n1+n2 = 99*11 + 1

similarly for N=6 , we have 999*111 + 1
N=8 , 9999*1111 + 1
so on...

Mine quite straightforward realization works about 1 sec. and uses almost 60 megabytes of memory. Please, give an idea how to solve it more quickly...

Input french string "Administrateurs" and compaire it with english string "Administrators".

Expected result:
True

Sample code:
string a = Environment.UserName;

if (a.Equals("Administrator",StringComparison.
CurrentCultureIgnoreCase))
{
    Console.WriteLine("Good");
}

output: Good

Anyone with the same WA? I think Java is starting to be the problem...

Is there a trick? I must forget something, because i tried 3 different implementations for finding the current convex hull of the points, and all three get WA 10...

Test
2.0 9.5 12
Ans
0

:)

Question removed

Edited by author 15.03.2010 03:09

что это за ошибка
может кто сказать что она значит

Hi,

I got WA at test #3. I have applied all of the samples above to the algorithm and the program output the correct result. Any body can help me?

i divide n numbers into ceil[n/a] segments
for each segment, increasing B's ties by add b until it reach a, and count the pairs.

such like this:
while(j + b <= a) {
    ans += j * b;
    j += b;
}

I'm from China.

If there is cycle in graph then -1,
if path from S to T does not exist then -1,
else 1.

I wonder why the solution with Arithmetic progression is impossible.
a:=a+b
c:=c-d
...
...
and several times
isn't that equals to a:=a+bx
and c:=c-dx ??

Mods, send input data for test1 please. Trying to solve the problem second day already. Thx

vanvector@mail.ru

Why the Sample Output is 3?
BIG is wrong, yes? It must be 6!
I don't understand the meaning!

What's wrong in my program? I can't find mistake. Anybody knows it?
#include<fstream>
#include<iostream>
#include<cstdio>
using namespace std;

    long n, m;
    __int64 x, y = 0, n1;

int main (){
    cin >> n >> m;
    n1 = 3*n;
    for (int i = 0; i < m; ++i){
        cin >> x;
        y+=x;
        if (y >= n1){
            cout << "Free after " << i+1 << " times.";
            return 0;
        }
    }
    cout << "Team.GOV!";
    return 0;
}

Is there somewhere a list with the restricted functions for Java?

B can be greater than A
Good Luck ^^