Soluție HackerRank pentru Divisibility of Power, subdomeniul Number Theory, în C++14. Include cerința formatată, exemple, explicația pașilor și cod sursă.
- Problemă: Divisibility of Power
- Domeniu: Number Theory
- Limbaj: C++14
Challenge: Divisibility of Power
Subdomeniu: Number Theory (number-theory)
Scor cont: 60.0 / 60
Submission status: Accepted
Submission score: 1.0
Submission ID: 464742999
Limbaj: cpp14
Link challenge: https://www.hackerrank.com/challenges/divisibility-of-power/problem
Cerință
You are given an array A of size N. You are asked to answer Q queries.
Each query is of the form :
text{i j x}
You need to print `Yes` if x divides the value returned from find(i,j) function, otherwise print `No`.
find(int i,int j)
{
if(i>j) return 1;
ans = pow(A[i],find(i+1,j))
return ans
}
Input Format
First line of the input contains N. Next line contains N space separated numbers. The line, thereafter, contains Q , the number of queries to follow. Each of the next Q lines contains three positive integer i, j and x.
Output Format
For each query display `Yes` or `No` as explained above.
Constraints
2 ≤ N ≤ 2 × 10^5
2 ≤ Q ≤ 3 × 10^5
1 ≤ i,j ≤ N
i ≤ j
1 ≤ x ≤ 10^16
0 ≤ value of array element ≤ 10^16
No 2 consecutive entries in the array will be zero.
Sample Input
4
2 3 4 5
2
1 2 4
1 3 7
Sample Output
Yes
No
Cod sursă
//divisibility-of-power.cpp
//Divisibility of Power
//Ad Infinitum - Math Programming Contest August'14
//Author: derekhh
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
long long mulmod(long long a, long long b, long long m)
{
long long res = 0;
while (a != 0)
{
if (a & 1) res = (res + b) % m;
a >>= 1;
b = (b << 1) % m;
}
return res;
}
long long ModExp(long long a, long long b, long long n)
{
long long c = 1, d = a;
while (b)
{
if (b & 1) c = mulmod(c, d, n);
d = mulmod(d, d, n);
b >>= 1;
}
return c % n;
}
long long a[300001];
int foo(long long a, long long b)
{
if (a < 2) return (int)a;
if (b == 0) return 1;
if (b == 1)
{
if (a > 64) return 64;
return (int) a;
}
if (a > 8 || b > 6) return 64;
int ret = (int)pow(a, b);
return ret > 64 ? 64 : ret;
}
int main()
{
long long n;
scanf("%lld", &n);
for (int i = 1; i <= n; i++)
scanf("%lld", &a[i]);
long long q;
scanf("%lld", &q);
while (q--)
{
long long i, j, x;
scanf("%lld%lld%lld", &i, &j, &x);
int res = 1;
int k = min(j, i + 6);
if (i + 7 <= j && a[i + 7] == 0) k--;
for (; k > i; k--)
res = foo(a[k], res);
long long ans = ModExp(a[i], res, x);
if (ans == 0) printf("Yes\n");
else printf("No\n");
}
return 0;
}