#include<iostream>
using namespace std;

int main()
{
    int x, y;
    int sum1 = 0, sum2 = 0;
    char n = 'n';

    cin >> x;
    int max = x + 1;
    int min = x - 1;
    for (int i = 0; i <3; i++)
    {
        y = max % 10;

        max = max / 10;
        sum1 += y;
    }
    for (int i = 0; i < 3; i++)
    {
        y = max % 10;

        max = max / 10;
        sum2 += y;

    }
    if (sum2 == sum1)
    {
        n = 'y';

    }

    y = 0;
    sum1 = 0;
    sum2 = 0;
    for (int i = 0; i <3; i++)
    {
        y = min % 10;

        min = min / 10;
        sum1 += y;
    }
    for (int i = 0; i <3; i++)
    {
        y = min % 10;

        min = min / 10;
        sum2 += y;
    }
    if (sum2 == sum1)
    {
        n = 'y';

    }
    if (n=='n')
    {
        cout << "no";
    }
    else
    {
        cout << "yes";
    }
    return 0;
}

choose k numbers k1,k2...km such that sigma(cos((2*ki+1)*pi/n))==0&&sigma(sin((2*ki+1)*pi/n))==0

that is  sigma(cos((2*ki+1)*pi/n)+sin((2*ki+1)*pi/n)*i)==0
because cos(k*theta)+i*sin(k*theta)==(cos(theta)+i*sin(theta)^k

so it is sigma(cos(pi/n)+i*sin(pi/n))^(2*ki+1)==0

I think this problem must involved this...

What is test # 14? Can anyone help me?


Edited by author 22.11.2017 20:14

#include<stdio.h>
#include<math.h>
#define n 4
int main(){
int a[n]={100000000};
int i;

for(i=0; i<n; i++)
scanf("%d", &a[i]);

for(i=n; i>0; i--){
   printf("%.4f\n", sqrt(a[i]));}

return 0;
}

#include<stdio.h>
#include<math.h>
#define n 4
int main(){
int a[n]={100000000};
int i;

for(i=0; i<n; i++)
scanf("%d", &a[i]);

for(i=0; i<n; i++){
   printf("%.4f\n", sqrt(a[i]));}

return 0;
}

Can someone know the 9th test?

Wrong answer 4. HELP!!!!

Edited by author 03.06.2016 17:42

Edited by author 03.06.2016 17:42

same code visul C++ 2017 0.249s
G++7.1  2.5s

Hi, everyone!
I used Manacher's algorithm to solve this problem, but I got WA on #14 test. Can someone hint me, what could be the problem?

please give the code to java
please give the code to java

Edited by author 18.11.2017 07:25

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{

        int n, a = 1, counter = 0;
        string s;
        cin >> n;

        for (int i = 0; i < n; ++i)
         {
          cin >> s;
                if (s[0] == 'A' || s[0] == 'P' ||s[0] == 'O' ||s[0] == 'R' &&  a == 2){ a = 1; counter++;}
            else
                 if (s[0] == 'A' || s[0] == 'P' ||s[0] == 'O' ||s[0] == 'R' &&  a == 3){ a = 1; counter +=2;}

                 if (s[0] == 'B' || s[0] == 'M' ||s[0] == 'S' &&  a == 1){ a = 2; counter++;}

            else

                 if (s[0] == 'B' || s[0] == 'M' ||s[0] == 'S' &&  a == 3){ a = 2; counter++;}


                if (s[0] == 'D' || s[0] == 'G' ||s[0] == 'K' ||s[0] == 'T' || s[0] == 'W' &&  a == 1){ a = 3; counter +=2;}

            else

                if (s[0] == 'D' || s[0] == 'G' ||s[0] == 'K' ||s[0] == 'T' || s[0] == 'W' &&  a == 2){ a = 3; counter++;}



        }

        cout << counter;
        return 0;

}

initially  let result sequence to be 10*  10* 10*   0* means zero or more zeroes.

every time choose two number and if we merge them produce a smaller number then we merge them.

the procedure stops until we can't merge to get a smaller number

is it correct?? seems to be..

Edited by author 16.11.2017 07:10

Anyone crash test#2

aaaaa
aaa

No
aaa

aaaaa
aaaaaaaaa

No
aaaa aaaaa

aabaa
aabaabaabaaaaabaa

No
aab aab aaba aa aabaa

aabaa
aabaabaab

No
aab aab aab

abcabcabccs
abcaabcaabcaa

No
abca abca abca a

aabaa
aabaabaaaa

No
aab aabaa aa

What's wrong? I have 'time limit excedeed' in 9th test.



import functools

def pr(n):
    result = []
    i = 2
    now = 1
    while n >= 1:
        if n % i == 0:
            now *= i
            n //= i
        elif n % i != 0 and now > 1:
            result.append(now)
            i += 1
            now = 1
        else:
            i += 1
        if n == 1:
            result.append(now)
            return result

def f(n):
    result = 0
    i = 1
    while i ** 2 < n:
        if n % i == 0:
            result += 1
        i += 1
    if i ** 2 == n:
        return 2 * result + 1
    else:
        return 2 * result

newList = [int(input()) for _ in range(10)]
product = functools.reduce((lambda x, y: x * y), newList)
print(functools.reduce(lambda x, y: x * y, map(f, pr(product))) % 10)

You don't must develop hard solution. Your work should not payed many time. Just pay attention, and look again on tests. Attention will save your time.
I am sorry, for my poor English. Good luck!

Edited by author 18.10.2015 16:52

Edited by author 11.11.2017 22:53

Edited by author 11.11.2017 22:53