# YVNUM Editorial

Author and Editorialist: Hussain Kara Fallah
Tester: Michael Nematollahi

NONE

### PROBLEM:

Given a decimal integer N (with no zeroes at all even as intermediate digits). Consider all left cyclic shifts of N. The left cyclic shift is the operation of removing the leftmost digit from the number and appending it to the end.

Suppose N = 123 then all left cyclic shifts are {123,231,312}. So you start with N and repeat the mentioned operation N-1 times.

Yalalovichik number of N is equal to the concatenation of all these cyclic shifts.

For the previous example:

Y = 123231312.

Given N, you need to compute the value of Y modulo 10^9+7

Length of N may reach 10^5

### EXPLANATION:

Letâ€™s assume that length of N is equal to L. We know that:

N = a_0*10^{L-1} + a_1*10^{L-2} + a_2*10^{L-3} + ... + a_{L-1}

Assuming that digits are indexed from 0 to L-1 and from left to right. a_0 represents the leftmost digit. So letâ€™s assume we need to do one shift operation, how to erase the leftmost digit?

N' = N - a_0*10^{L-1}

N' is equal to the value of N after removing the leftmost digit.

Letâ€™s assume that the number resulting from the shift is M so:

M = N' * 10 + a_0

so now the number resulting from the shift is M we need to append it to our final answer. At the beginning the answer ans = N. To append it letâ€™s shift ans to the left by L digits and simply add M.

ans = ans * 10^L + M

This is for the first shifting operation, to solve the problem repeat this operation N-1 times, but donâ€™t forget to continue working on the shifted number.

We only need to calculate 10^L and 10^{L-1}, this can be done in linear time or faster if you prefer.

To maintain the answer modulo 10^9+7 we calculate the powers mentioned above modulo 10^9+7 and we always take our results modulo 10^9+7 after each subtraction or multiplication or addition.

Refer to the implementations for more details.

Total Complexity O(L)

### AUTHORâ€™S AND TESTERâ€™S SOLUTIONS:

AUTHORâ€™s solution

TESTERâ€™s solution

1 Like

Thanks, very simple and good editorial.

Why did you do cur+=MOD ?

Thanks, for this easy explanation . i am constantly getting TLE for this but now i can solve this

i just want to know that is the constraint that N<=10^5 for integer or string of length 10^5

1 Like

The link to my solution:

Can anybody help? Simple solution but doesnâ€™t work: https://ideone.com/nEqSN1

I was getting TLE as i was doing in O(L*L),now i can do it,good editorial.

It could be rewritten M = N * 10 - a_0 * (10^L - 1). So we only need to calculate 10^L.

And we can do it while calculating N.

I can understand the explanation and logic pretty wellâ€¦but I am not getting anything seeing the solutionâ€¦ :disappointed_relieved

1 Like

Nice Editorial

have you been able to understand the reason??? I understand that it is to avoid negative number in the previous operation but how that happens I would also like to know.Please help us @watcherâ€¦

Hello,@itz_pankaj can you explain to me the cur+=MOD operation as well cur%=MOD operation??I get that these operations are compensating any negative result that may have been produced in previous step but not able to understand how it is being achieved.

my code is correct but i am getting wrong answer can anyone please point out my mistake.

#include<bits/stdc++.h>
using namespace std;
long long int m=1000000007;
long long int power(int x, int y)
{
long long int res = 1;

``````while (y > 0)
{

if (y & 1)
res = (res*x)%m;

y = y>>1;
x = ((x*x)%m);
}
return ((res)%m);
``````

}
int main()
{
long long int t,i,j,d,k,n;
cin>>t;
for(i=0;i<t;i++)
{
cin>>n;
n=abs(n);
int c=0;
int x=n;
while(x!=0)
{
c++;
x=x/10;
}
x=n;
int a[c],c1=0;
while(x!=0)
{
d=x%10;
a[c1]=d;
c1++;
x=x/10;
}
x=power(10,c);
long long int ans=n;
long long int nn,mm;
for(j=1;j<c;j++)
{
nn=(n-(a[c-j](power(10,c-1))%m))%m;
nn=((nn
10)%m+a[c-j])%m;
cout<<nn<<endl;
ans=((ans*x)%m+nn)%m;
n=nn%m;
ans=ans%m;
}
cout<<ans<<endl;
}
return 0;
}

I will see your codeâ€¦but I would advise you for this problem that you first think out the algo without thinking about Modulus.Then after you have done soâ€¦apply modulus at each stage of any operation
you do.I hope you will get AC