# Can anyone help ccdsap mock question?

can anyone propose a solution for this i have tried backtracking for it but getting tle?
Can anyone give me a better approach?

Basically go from left to right, calculate the number of ways of getting each remainder, starting with ans/dp[0][0]=1;
ans[i][j] represents number of ways to get j mod m using this first i digits(left to right)
also
p2[i] =2^i mod m
After that we know that we can calculate the dp for the next digit by doing this if it is a 1

``````ans[i+1][(j+p2[l-i-1])%m]+=ans[i][j];
``````

If it is a 0 we do

``````ans[i+1][j]+=ans[i][j];
``````

for β_β we do both as it can be either
This solves it in O(nm).
Donβt know why the constraints are so low
https://www.codechef.com/viewsolution/29878828

``````#include <iostream>
#include <bits/stdc++.h>
#include <cmath>
#include <vector>
#define ll long long int
#define mp make_pair
#define pb push_back
#define vi vector<int>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >>t;
while(t--){
int l,m;
cin>>l>>m;
int p2[l]; //power of 2 mod m
p2[0]=1;
for(int i=1;i<l;i++){
p2[i]=(2*p2[i-1])%m;
}
ll ans[l+1][m];// number of numbers j mod p possible with the first i digits
for(int i=0;i<=l;i++){
for(int j=0;j<m;j++){
ans[i][j]=0;
}
}
ans[0][0]=1;
for(int i=0;i<l;i++){
char a; //because we want one digit at a time
cin>>a;
if(a=='1'){
for(int j=0;j<m;j++){
ans[i+1][(j+p2[l-i-1])%m]+=ans[i][j];
}
}
else if(a=='0'){
for(int j=0;j<m;j++){
ans[i+1][j]+=ans[i][j];
}
}
else{
for(int j=0;j<m;j++){
ans[i+1][j]+=ans[i][j];
ans[i+1][(j+p2[l-i-1])%m]+=ans[i][j];
}
}
}
cout<<ans[l][0]<<"\n";

}

}
``````
4 Likes

huh!
when will i be able to think so profoundly!
amazing.Thanks