BigMod and BigsumMod

/* Bismillahir Rahmanir Rahim BigMod of Series: (1 + b + b^2 + b^3 + b^4 +......+ b^(p-1) )%mod */ #include<bits/stdc++.h> #define ll long long using namespace std; ll b, p, m; ll bigmod(ll b, ll p, ll m){ // its the BigMod if(p==0) return 1; ll x=bigmod(b, p/2, m); x=(x*x)%m; if(p%2==1) x=(x*b)%m; return x; } ll bigsummod(ll b, ll p, ll m){ // mod in every line if(p==2) return (1+b)%m; // when 2 term (1+b) else if(p==3) return (1+b+b*b)%m; else if(p%2==1) return (1+b*bigsummod(b, p-1, m))%m; else{ ll xx=bigsummod(b, p/2, m)%m; return xx=(xx+(bigmod(b, p/2, m)*xx))%m; } } int main(){ while(cin>>b>>p>>m){ // b=Base, p=power+1(number of term), m=mod cout<<bigsummod(b, p, m)<<endl; } return 0; }

Solution of UVa-11029 Leading and Trailing

#include<bits/stdc++.h>

#define ll long long

using namespace std;

ll n, k, t_case;

ll bigmod(ll b, ll p, ll m){

    if(p==0)return 1;

    ll xx=bigmod(b, p/2, 1000);

    xx=(xx*xx)%1000;

    if(p%2==1)xx=(xx*b)%1000;

    return xx;
}

int main(){

    cin>>t_case;

    while(t_case){

        cin>>n>>k;

        /* executing first 3 digits */

        double x=k*(log10(n));

        x=x-(int)x; // taking fraction value only
        

        double ans=pow(10, x);

        ans=ans*100;

        cout<<(int)ans<<"...";

        /* executing last 3 digits */

        printf("%03d\n", bigmod(n, k, 1000));

        t_case--;

    }

    return 0;
}