HackerEarth Maximize the modulo function problem solution

In this HackerEarth Maximize the modulo function problem solution You are given an integer N that is represented in the form of string S of length M. You can remove at most 1 digit from the number after removing the rest of the digits that are arranged in the same order.

HackerEarth Maximize the modulo function problem solution.

#include<bits/stdc++.h>
#define int long long int
using namespace std;

int power(int a, int b, int k)
{
    if(b == 0) return 1;
    int ans = 1;
    int val = a;
    while(b)
    {
        if(b%2)
        {
            ans *= val;
            ans %= k;
        }
        val *= val;
        val %= k;
        b /= 2;
    }
    return ans;
}

void solve(){
    int m, k;
    cin >> m >> k;

    string s;
    cin >> s;

    int mod_val = 0;
    for(int i = 0 ; i < m ; i++)
    {
        int digit = s[i] - 48;
        mod_val = (mod_val + (digit*power(10, m - i - 1, k)%k))%k;
    }

    int prev = 0;
    int answer = mod_val;
    for(int i = 0 ; i < m ; i++)
    {
        int digit = s[i] - 48;
        mod_val = (mod_val - (digit*power(10, m - i - 1, k))%k + k)%k;
        answer = max(answer, (prev + mod_val)%k);
        prev = (prev + (digit*power(10, m - i - 2, k))%k)%k;
    }

    cout << answer << endl;
}

signed main(){
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}

Second solution

t = int(input())
while t > 0:
    t -= 1
    m, k = map(int, input().split())
    n = input()
    pre = [0]
    for c in n:
        pre += [(pre[-1] * 10 + int(c)) % k]
    ans = pre[-1]
    suf = 0
    cur_pow = 1
    for i in range(m - 1, -1, -1):
        ans = max(ans, (pre[i] * cur_pow + suf) % k)
        suf += cur_pow * int(n[i])
        suf %= k
        cur_pow = cur_pow * 10 % k
    print(ans)

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