Can you Solve!

help
1

How to solve this problem using dp when n<=10^8
can any on explain???

2

Do you know about this concept? If you don’t go through this, you will get this question easily.
In case you don’t, I will write it’s program and send it here, as soon as possible.

Using dp it can be done as this, but it will not pass all test cases.

    /*author - Ashutosh Chitranshi*/
    #include<bits/stdc++.h>
    using namespace std;
    #pragma GCC push_options
    #pragma GCC optimize ("unroll-loops")
    #define watch(x) cout << (#x) << " is " << (x) << "\n"
    #define watch2(x,y) cout <<(#x)<<" is "<<(x)<<" and "<<(#y)<<" is "<<(y)<<"\n"
    #define pow2(x) ((x)*(x))
    #define ll long long
    #define ld long double
    #define eb emplace_back
    #define pb push_back
    #define pf push_front
    #define mod 1000000007
    #define clock (clock() * 1000.0 / CLOCKS_PER_SEC)
    #define mp make_pair
    #define ff first
    #define ss second
    #define null NULL
    #define all(c) (c).begin(),(c).end()
    #define nl "\n"
    typedef vector<int> vi;
    typedef vector<ll> vl;
    typedef vector< vi > vvi;
    typedef vector< vl > vvl;
    typedef pair< int,int > ii;
    typedef pair< ll,ll> pll;
    typedef map< ll,ll> mll;
    typedef map< int,int> mii;
    int main()
    {
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        ll t;
        cin >> t;
        while(t--)
        {
            ll n;
            cin >> n;
            ll dp[n+1];
            dp[0] = 9;
            dp[1] = 31;
            for(int i=2;i<=n;i++)
            {
                dp[i] = ((8*dp[i-1])%mod - (15*dp[i-2])%mod + mod)%mod;
            }
            cout << dp[n] << "\n";   
        }
        return 0;
    }
3

yeah thanks i have tried by understanding of that topic but i got wrong
could you tell where i am wrong??
my solution link

4

Your submission is not visible.

5

you can try something like this

6

Can anyone explain what’s wrong here? Forget about internal tests cases, its printing 2nd example output more than 100 times.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {

    long long int tests;
    cin >> tests;
    while(tests--){
        unsigned long long int n;
        cin >> n;
        if(n==1) cout << 31 << "\n";
        else {
            unsigned long long int f[n];
            long long int m=1000000007;
            f[0]=9;
            f[1]=31;
            long long int k=2;
            while(k<=n){
                f[k]=8*f[k-1]%m-15*f[k-2]%m;
                k++;
            }
            cout << f[n] << "\n";
        }
    }
    return 0;
}
7

thank you i have figured it out by correcting (a-b)%m formula and it is accepted :upside_down_face: :hugs: