I get 100/100 point :)

Solution:

• Call it as Ab (e.g: if 2345 -> A = 234 and b = 5), and Ab have n length

• We need to resolve: Ab * x = bA

  => A = {b[10^(n-1) - x]} / (10x - 1)

  (I will explain later)

  with:

  - x = 1 to 9
  
  - n = 2 to m
  
  - b = 1 to 9
  
  - length of A is equal to (n-1)
  

• We can find A with above conditions. And in Java, we use BigInteger to calculate.

explain:

Ab * x = bA
   
=>   (10A + b) * x = b * (10^(n-1)) + A 
   
=>   10x * A + b * x = b * (10^(n-1)) + A

=>   (10x - 1) A = b[ 10^(n-1) - x]

=>   A = {b[10^(n-1) - x]} / (10x - 1)

This is not clear by last 5 digits...

#include<stdio.h>
int a[105]={0,0,495,5490,55485,55480,98331,98326,98321,98316,98311,98306,41157,75329,75324,75319,75314,75309,28684,28679,28674,28669,19968,19963,62814,62809,96981,96976,90073,90068,32919,32914,32909,32904,32899,32894,86269,86264,86259,20431,20426,20421,34700,34695,50713,50708,50703,50698,93549,93544,93539,93534,27706,27701,81076,81071,74168,74163,63988,63983,6834,6829,6824,6819,6814,40986,75141,75136,75131,75126,75121,75116,28491,28486,28481,28476,28471,28466,5494,5489,5484,5479,5474,5469,12850,12845,12840,12835,28853,28848,82223,16395,16390,16385,16380,16375,59226,59221,59216,59211,59206};
int main (){
	int m;
	scanf ("%d",&m);
	printf ("%d\n",a[m]);
	return 0;
}

I made a list and successfully passed the tests. You can check the correctness of your code through the list.

What is test-case 7? Only cannot get it right, weird...

It says m < 100. So, for very large value of m say m=99, n will belong to values between 10 and 1e99. How are you guys dealing with the bigInteger?

For me, some easy test cases are passing but others are giving time out!

i thought i understood hte problem, but hte sample input 0/output 0 confuse me

n is between 10, 100

how is 495 a possible right rotation of those nums ?

n = int(input())

sum = 0

for i in range(2, n+1):

    for j in range(1, 9+1):
		
        for k in range(1, 9+1):
				
            temp = (j*(10**(i-1) - k))%(10*k - 1)
						
            if temp == 0:
						
                ans = f"{int((j*(10**(i-1) - k))/(10*k - 1))}"+ f"{j}"
								
                if len(ans) > 5:
								
                    sum += int(ans[-5:len(ans)+1])
                    print(ans)
										
                else:
                    sum += int(ans)
                    print(ans)

Please tell me what I am doing wrong, this code is only passing 2 test cases.       

If 495 is the result of the scenario when n=2 and the constituent numbers are 11,22,33,44,55,66,77,88,99 then... How are 22,33,44,55,66,77,88,99 special numbers? Obviously 11 is... (1 * 10) + 1 = 1 * (1 * 10) + 1 But take 22 for instance: 22 is (2 * 10) + 2 != 2 * (2 * 10) + 2 By my understanding 22 is not a special number, neither are 33,44,...,99.

1}

include

include

int ff(long long int a) { int c=0; while (a>0) { a=a/10; c=c+1; } return c; } long long int gg(long long int a) { long long int p,q,s; q=ff(a); p=a%10; a=a/10; s=pow(10,q-1)*p+a; return s; } int main() { long long int i,j,k,s=0; long long int n; scanf("%lld",&n); for (i=11;ik) break; } } printf("%lld",s); }

2}

def dov(a): p=str(a) q=p[len(p)-1] r=p[0:len(p)-1] s=q+r; return int(s) a=int(input()) s=0 for x in range(11,10**a): p=dov(x) for y in range(1,10): if (p==x*y): s=s+x elif x*y>p: break
print(s)

can anyone tell me why both process terminated

its kind of strange,

I am using the below code m=int(input()) m=6 sumd=0 i=11 while 10 answer=str(sumd) answer=int(answer[-5:]) print(answer)

in python, when m=6, I can retrieve 142857 but m=2 doesnt work, I only get 11