Nyoj 1197--will you add it? —————— "Fast power, divide and conquer"

Will you add it?  Time limit: Ms | Memory limit: 65535 KB Difficulty: 3

Describe

Give two integers a and N, calculate (A + a^2 + a^3 + ... + a^ (N-1) + a^n)% 666666.

Input

multiple sets of test data.
Each set of data contains two integer a,n (0≤a,n≤10^18).

Output

output (A + a^2 + a^3 + ... + a^ (N-1) + a^n)% 666666 is how much, each group of data occupies one row.

Sample input

2 510 20

Sample output

 62110 
 
 
 
  We can always become: when N%2==1, (A1+A2+A3...+AN/2) +an/2* (a1+a2+a3+ ... AN/2) +an. 
 is (A1+A2+A3...+AN/2) * (an/2+1) +an. 
 when N%2==0, (A1+A2+A3...+AN/2) +an/2* (a1+a2+a3+ ... AN/2). 
 is (A1+A2+A3...+AN/2) * (an/2+1). The 
 can then be divided into solutions. 
 
 
  

#include <bits/stdc++.h>using namespace std; #define LL Long Longconst ll mod=666666; ll Pow (ll a,ll x) {  //fast power    ll ret=1;    while (x) {        if (x&1)          ret=ret*a%mod;        A=a*a%mod;        x>>=1;    }    return ret;} ll CALCU (ll a,ll N) {    //Sub    -rule if (n==1)        return A;    LL RET=CALCU (A,N/2);//The return value    ret=ret* (1+pow (A,N/2))%mod of the split part;    if (n%2==1)      //n If an odd number is required plus a^n        ret= (Ret+pow (a,n))%mod;    return ret;} int main () {    LL a,n;    while (scanf ("%lld%lld", &a,&n)!=eof) {        LL ans= calcu (a%mod,n);        printf ("%lld\n", ans);    }    return 0;}

　　

Nyoj 1197--will you add it? —————— "Fast power, divide and conquer"