Let C  = A + '$*&' + B
for string C build Palindrom Tree.
calculate frequency of each palindrom : f(A,p)
calculate frequency of each palindrom : f(B,p)
x = count(f(A,p)>f(B,p))
y = count(f(A,p)==f(B,p) && f(A,p)!=0)
z = count(f(A,p)<f(B,p))

//little code:  freq[i][0] ---> f(A,p);   freq[i][1] --> f(B,p).
for(int i = 0; i < a.size(); ++i){ add_letter(a[i]); freq[last][0]++;}

add_letter('$'); add_letter('*'); add_letter('&');

for(int i = 0; i < b.size(); ++i){ add_letter(b[i]); freq[last][1]++;}

for(int z = sz-1; z > 0; z--){
    v = link[z];
     freq[v][0] += freq[i][0];
     freq[v][1] += freq[i][1];
}
int x=0,y=0,z=0;
for(int i = 2; i< sz; ++i)// skip roots
{
    x += (bool)(freq[i][0] > freq[i][1]);
    y += (bool)(freq[i][0] == freq[i][1] && freq[i][0] !=0);
    z += (bool)(freq[i][0] < freq[i][1]);
}

#include <iostream>
#include <string>
using namespace std;
int main() {
    int num, res;
    string c, b = "!", a = "!";

    cin >> num >> c;
    res = num;
    int sign = 0;
    for (int i = 1; b <= c; i++) {
        b = b + a;
        sign = i;
    }
    if (num%sign != 0) {
        for (int i = 1; i*sign < num; i++) {
            res = res * (num - i*sign);
        }
        res = res *(num%sign);
    }
    else {
        for (int i = 1; i*sign < num; i++) {
            res = res * (num - i*sign);
        }
    }
    cout << res;

    return 0;
}

Solved both tasks with the same algorithm. Think this task is over-valued.

Edited by author 26.10.2016 20:59

just submitted code and got TLE on 15test, submitted same code after few minutes and got accepted. submitted again and got TLE on 24test for same code.

All solutions have been rejudged with the current time limit. Also, few more tests have been added. 10 authors have got AC while 23 other have lost.

RT

source code of algorithm Manaker (find all palindroms in O(n)), where explained in e-maxx.ru is incorrect :), see comments to below of that page.

Edited by author 25.10.2016 16:17

it is also not worth than 10000+ rating...

1 5
3 2 1 1 2

RT

I'm getting wrong ans on test case 3. Can someone please give a hint?

Here's my solution in Java:

FastReader sc = new FastReader(System.in);
        double a = sc.nextDouble();
        double r = sc.nextDouble();
        double ans;

        if (r <= a / 2) {
            ans = Math.PI * r * r;
        } else {
            double angle = (2 * Math.PI) - (8 * Math.acos(a / (2 * r)));
            ans = (r * r * angle / 2) + (2 * a * Math.sqrt(r * r - a * a / 4));
        }

        System.out.println(String.format("%.3f", ans));

New tests have been added to the problem, all accepted solutions have been rejudged. 490 authors have lost their AC.

Can anyone provide test case for WA#1, please?

#include <iostream>
#include <cstdlib>
#include <string>
using namespace std;

string S, A;
unsigned short An = 1;
bool negative     = true;

void get_next_A();
void get_S(unsigned short);

int main()
{
    unsigned short input_n;

    cin >> input_n;

    get_S(input_n);

    cout << S << endl;

    return 0;
}

void get_next_A()
{
    if(An == 1)
        A = "sin(1)";
    else
    {
        A.erase(A.size() - An + 1, An);

        if(negative)
            A += "-sin(";
        else
            A += "+sin(";

        char buffer[5];

        itoa(An, buffer, 10);
        A += buffer;

        for(unsigned short j = 0; j < An; ++j)
            A += ")";

        negative = !negative;

    }

    ++An;
    return;
}

void get_S(unsigned short n)
{
    char _buffer[10];

    for(unsigned short i = 0; i < n - 1; ++i)
        S += '(';

    for(unsigned short i = n; i > 0; --i)
    {
        get_next_A();

        itoa(i, _buffer, 10);

        S += A;
        S += '+';
        S += _buffer;
        if(i != 1)
            S += ')';

    }

    return;
}

#include <iostream>
using namespace std;
int main() {
    int n,res=0;
    cin >> n;
    if (n < 0) {
        n = n*-1;
        int *arr;
        arr = new int[n+2];
        arr[0] = 1;
        for (int i = 1; i < n+2; i++) {
            arr[i] = arr[i - 1] - 1;
        }
        for (int i = 0; i < n+2; i++) {
            res = res + arr[i];
        }
        cout << res;
    }
    else if(){
        int *arr;
        arr = new int[n];
        arr[0] = 1;
        for (int i = 1; i < n; i++) {
            arr[i] = arr[i - 1] + 1;
        }
        for (int i = 0; i < n; i++) {
            res = res + arr[i];
        }
        cout << res;
    }
    return 0;
}

program Suma;
var a,b,e:integer;
begin
read(a);
b:=1;
if a<0 then
e:=((1+a) div 2)*(((a-1)div((b-1)-1))+1)
else
e:=(a*(a+1)) div 2;
writeln(e);
end.

var i,n,q,h:integer;
k:real;
begin
q:=0;
read(n);
h:= 10000;
if (n <= h) and (n >= -h) then begin
if n>0 then begin
k:= ((1+n)/2)*n;
writeln(k);
end;
if n=0 then begin
k:=0+1;
writeln(k);
end;

if n<0 then begin
for i:= 1 downto n do
inc(q);
k:= ((1+n)/2)*q;
writeln (k);
end;
end;
end.

I tried binary search, harsh table, got TLE. then I thought using direct mapping table could get MLE, but never TLE, so i tried it, still TLE!!!tell me how could this code get TLE.
#include <iostream>
using namespace std;
char a[1000000001];
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int j;
        cin >> j;
        a[j] = 1;
    }
    int m;
    cin >> m;
    int cnt= 0;
    for (int i = 1; i <= m; ++i) {
        int j;
        cin >> j;
        cnt += a[j];
    }
    cout << cnt;
}