 # SUB7NUM-Editorial

Setter: Yash Kulkarni
Tester: Manan Grover, Lavish Gupta
Editorialist: Devendra Singh

1943

# PREREQUISITES:

Dynamic programming

# PROBLEM:

Kulyash has given you an array A of size N.

He defines the subsequence-number of a non-empty subsequence
S of array A as the number formed by the concatenation of all the elements of the subsequence S.

Find the count of non-empty subsequences of A having their subsequence-numbers divisible by 7. Since the answer can be huge, output it modulo 10^9 + 7.

For example: Consider A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]. A subsequence S of A is [2, 5, 7, 10]. The subsequence-number of this subsequence is 25710.

# QUICK EXPLANATION:

Let dp[i][j] represent the number of subsequences in the prefix P_i of the array A which leave a remainder j when there subsequence-number is divided by 7.
Let D be the number of digits in the i^{th} element.
Initialize dp=1 and rest as 0 (the empty subsequence, this will be subtracted from the answer later).
The dp transition is : for each j from 0 to 6
dp[i][j*10^{D} + A_i] = dp[i-1][j] + dp[i-1][j*10^{D}+A_i]

# EXPLANATION:

For each value of remainder, rem = 0\: to\: 6, let us calculate the number of subsequences in the prefix P_i (first i elements of the array A) which leave a remainder rem when their subsequence-number is divided by 7 . The subsequences till a particular index i leaving remainder rem include the subsequences calculated for prefix P_{i-1} for the same remainder rem.
Let D be the number of digits in the i^{th} element. Let the number of subsequences in P_{i-1} which leave a remainder R when their subsequence-number is divided by 7 be X, Then the remainder when the subsequence-number formed by combining the i^{th} element with an earlier subsequence from these X subsequences is divided by 7 is (R*10^D+A_i) %7. Hence, we need to add X to the number of subsequences in P_i which leave a remainder (R*10^D+A_i)%7 when their subsequence-number is divided by 7.

This problem has both optimal substructure property and overlapping subproblems. These kind of problems can be solved using dynamic programming.

Let dp[i][j] represent the number of subsequences in the prefix P_i of the array A which leave a remainder j when there subsequence-number is divided by 7.
Initialize dp=1 and rest as 0 (the empty subsequence, this will be subtracted from the answer later).
The dp transition is : for each j from 0 to 6
dp[i][j*10^{D} + A_i] = dp[i-1][j] + dp[i-1][j*10^{D}+A_i]

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

Setter's solution
``````#include<bits/stdc++.h>
#define ll long long
#define MOD 1000000007
using namespace std;
int main(){
ll T;
cin >> T;
while(T--){
ll n;
cin >> n;
vector<ll>v(n);
for(ll i=0;i<n;i++)cin >> v[i];
vector<vector<ll>>dp(n,vector<ll>(7,0));
vector<ll>mul(n,1);
for(ll i=0;i<n;i++){
ll x=v[i];
while(x){
mul[i]*=10;
x/=10;
}
}
dp[v%7]=1;
for(ll i=1;i<n;i++){
for(ll k=0;k<7;k++)dp[i][k]=dp[i-1][k];
for(ll k=0;k<7;k++)dp[i][(mul[i]*k+v[i])%7]=(dp[i][(mul[i]*k+v[i])%7]+dp[i-1][k])%MOD;
dp[i][v[i]%7]=(dp[i][v[i]%7]+1)%MOD;
}
cout << dp[n-1] << endl;
}
return 0;
}
``````
Editorialist's Solution
``````#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
int dig(int n)
{
int cnt = 0;
while (n)
{
cnt++;
n /= 10;
}
return cnt;
}
void sol(void)
{
int n;
cin >> n;
vll v(n);
for (int i = 0; i < n; i++)
cin >> v[i];
ll dp[n + 1], ans = 0;
memset(dp, 0, sizeof(dp));
dp = 1;
for (int i = 0; i < n; i++)
{
int digits = dig(v[i]), tenpower = pow(10, digits);
for (int j = 0; j < 7; j++)
{
dp[i + 1][(j * tenpower + v[i]) % 7] = dp[i][j] + dp[i][(j * tenpower + v[i]) % 7];
dp[i + 1][j] %= mod;
}
}
cout << (dp[n] - 1 + 2 * mod) % mod << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}

``````
3 Likes
``````lli h(lli arr[],int i,int n,int rem){
if(i==n){
return 0;
}
// if((((rem*10)%7+arr[i])%7)==0)cout<<rem<<" "<<i<<endl;
if(dp[rem][i]!=-1){
return (dp[rem][i])%MOD;
}
lli a=h(arr,i+1,n,rem)%MOD;
int num=woper(10,lennum(arr[i]));
lli b=(h(arr,i+1,n,(((rem*num)%7+arr[i])%7))%MOD+(((((rem*num)%7+arr[i])%7)==0)?1:0))%MOD;
dp[rem][i]=(a+b)%MOD;
return dp[rem][i];
}
``````

one with recursion

2 Likes

Thanks for the recursion part

nice problem @kulyash

1 Like

https://www.codechef.com/viewsolution/65695392
soln using memoisation

Thanks!!

In the transition: `dp[i][j*10^{D} + A_i] = dp[i-1][j] + dp[i-1][j*10^{D}+A_i]`, why is this term(`dp[i-1][j*10^{D}+A_i]`) added?