#include<stdio.h>
#include<string.h>
int f(int n,int m)
{
    if(n==1||m==1)
        return 0;
    if(n==2&&m>1)
        return 2;
    if(n==3&&m==2)
        return 3;
    if(m>2&&n>2)
    return f(n-2,m-2)+4;

}
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    printf("%d",f(n,m));
    fflush(stdin);
    getchar();
    return 0;
}

* The minimal height of the wall is set to 1 (previously 0).
* The total number of bricks is set to be between 1 and 100 000 (previously, not more than 1 000 000, but there were no such tests).
* New tests have been added.
* All accepted solution have been rejudged, 98 authors have lost their AC.

[code deleted]

Edited by moderator 19.11.2019 23:13

It looks as first input number (sequence length) isn't needed for solution.

int a[10]={-1,1,2,2,1,-1,-2,-2,-1,1};
...
for(i=1;i<=8;i++) if(x+a[i-1]>0 && x+a[i-1]<9 && y+a[i+1]>0 && y+a[i+1]<9) h++;

DO IT!!! :)

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?