If you use mask as two-dimensional array you probably will get TLE, instead of this use bitset from STL

Edited by author 20.09.2021 15:25

5
2 4 3 1 5

Not a proof

5
2 4 1 3 5

Cheater

1 0 1 0 0 0
11
ans = 10

n/=2;
dp[n][n*9];
dp[i][j] = dp[i-1][j]+dp[i-1][j-1] + ... + dp[i-1][j-9];
ans = dp[n][0]*dp[n][0] + ... + dp[n][n*9]*dp[n][n*9];

Edited by author 17.09.2021 19:12

Edited by author 17.09.2021 19:12

Precalculate pythagorean tripples

9
3 9 8 2
3 7 3 1
2 4 2
2 5 3
3 7 6 4
3 9 8 5
2 5 2
2 6 1
2 6 1
Please try this data,please
My progame can't pass this,but I got AC

My code:
n,m = map(int, input().split())
dct = dict()
for i in range(m):
    a = input()
    if a not in dct:
        dct[a] = 1
    else:
        dct[a] += 1
for i in dct:
    print("{0:.2f}%".format((dct[i]/m)*100))


Where is mistake?

Edited by author 13.09.2021 21:32

Hint: you should only know fundamental theorem of arithmetic

Треугольное число

The question stated that 0 < N < 30.  However, in test case #9 N is exactly 30, contradicting the constraint.

Посоветуйте пожалуйста сайты, где есть хорошая музыка в формате мп3 (не только русская), но также и зарубежная.

First I got WA#22 while I used double
than I change to use longdouble instead of double and I got WA#24
After that I change floor(x)/ceil(x) to floor(x+esp)/ceil(x-esp) and I got WA#24 too
But when I try to use double again I got AC!
[It's strange,too strange.I hate Accuracy issues]
(Sorry for my poor English XP)

Edited by author 07.09.2021 16:36

I guess that case is sth like a==0(a=-5Nz) ,b!=0 (b= VxNx+VyNy+VzNz)
(I find a check t=c/b instead of t=-b/c by mistake,and got wa)

there are 24 unique sequences of edges for the specific cube, i use 12 sequences and flip flop front with back for each of them
    static int R[][] = new int[][] {
        {1,2,3, 4,5, 6},
        {5,3,1, 4,2, 6},
        {2,1,5, 4,3, 6},
        {3,5,2, 4,1, 6},

        {6,4,3, 1,5, 2},
        {5,3,6, 1,4, 2},
        {4,6,5, 1,3, 2},
        {3,5,4, 1,6, 2},

        {1,2,6, 3,4, 5},
        {6,4,2, 3,1, 5},
        {2,1,4, 3,6, 5},
        {4,6,1, 3,2, 5}
    };
    static int front_back_rotate[][] = new int[][]{{0,1,2,3,4,5},{0,1,4,5,2,3}};
    //left right top  front bottom back
    //0       1     2    3     4      5
    static boolean eqv(int a[], int b[]) {
        for (int[] r: R) {
            for (int[] I: front_back_rotate) {
                int front_back[] = new int[6];
                for (int i=0; i<6; i++) {
                    front_back[r[I[i]]-1] = a[i];
                }
                if (Arrays.equals(front_back, b)) return true;
            }
        }
        return false;
    }

What is test 66?

Take attention:
No matter whose turn it is,if all the color is A,than Alise win,else Bob win.

Just like the title,I know how to solve this problem,becouse I have played the game before,and I know there is a way to move the chess.But if somebody dosen't play the game at all before,than how can he solve the problem?(Sorry for my poor English)

Edited by author 03.09.2021 16:00

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main() {

   int n, k;

   scanf("%d %d", &n, &k);

   if (n < 1 || n > pow(10, 9) || k < 1 || k > pow(10, 9)) {
      return 0;
   }

   unsigned long long result = 0;
   unsigned long long buffer = 1;
   n--;

   for (int i = 0; n >= 0; i++) {

      n -= buffer;
      result++;
      buffer *= 2;

      if (n <= 0) {
         printf("%llu\n", result);
         return 0;
      }

      if (buffer >= k) {
         buffer = k;
         double count = (double)n / buffer;
         if (count != (long long)count) {
            count++;
         }
         result += count;
         printf("%llu\n", result);
         return 0;
      }
   }
   return 0;
}

Edited by author 21.01.2020 01:57