Hackerrank--Divisibility of Power(Math)

You are given an array

Each query is of the form :

You need to print Yes if
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

Output Format

For each query display Yes or No as explained above.

Constraints

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

首先，对于每次询问（i，j，x）， 如果x中含有a[i]中没有的质因子，那么一定是No
其次，求出需要几个a[i]才能被x整除之后（设为cnt），就需要判断find(i+1, j)和cnt的大小。
对于a[i] = 0 或者 1 的情况可以进行特判，其他情况，因为x不大于1e16,所以可以直接暴力。
Accepted Code:

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <iostream>
  5 using namespace std;
  6 const int maxn = 200002;
  7 typedef long long LL;
  8 const int inf = 0x3f3f3f3f;
  9 LL a[maxn], x;
 10 int n, Q, i, j, cnt, near0[maxn], near1[maxn], c[maxn];
 11 bool flag;
 12 
 13 void init() {
 14     memset(near0, 0x3f, sizeof(near0));
 15     memset(near1, 0x3f, sizeof(near1));
 16     cnt = 0;
 17     for (i = 1; i <= n; i++) if (a[i] == 0) c[cnt++] = i;
 18     int be = 1;
 19     for (i = 0; i < cnt; i++) {
 20         for (j = be; j < c[i]; j++) near0[j] = c[i];
 21         be = c[i] + 1;
 22     }
 23     cnt = 0;
 24     for (i = 1; i <= n; i++) if (a[i] == 1) c[cnt++] = i;
 25     be = 1;
 26     for (i = 0; i < cnt; i++) {
 27         for (j = be; j < c[i]; j++) near1[j] = c[i];
 28         be = c[i] + 1;
 29     }
 30 }
 31 
 32 void read(LL &res) {
 33     res = 0;
 34     char c = ' ';
 35     while (c < '0' || c > '9') c = getchar();
 36     while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
 37 }
 38 void read(int &res) {
 39     res = 0;
 40     char c = ' ';
 41     while (c < '0' || c > '9') c = getchar();
 42     while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
 43 }
 44 
 45 LL gcd(LL a, LL b) {
 46     if (!b) return a;
 47     else return gcd(b, a % b);
 48 }
 49 
 50 bool ok(int be, int en) {
 51     LL res = 1;
 52     for (int i = en; i >= be; i--) {
 53         //if (quick_pow(a[i], res, res)) return true;
 54         LL tmp = 1;
 55         for (int j = 0; j < res; j++) {
 56             if (tmp >= cnt || a[i] >= cnt) return true;
 57             tmp *= a[i];
 58         }
 59         if (tmp >= cnt) return true;
 60         res = tmp;
 61     }
 62     return res >= cnt;
 63 }
 64 int main(void) {
 65     read(n);
 66     for (i = 1; i <= n; i++) read(a[i]);
 67     init();
 68     read(Q);
 69     while (Q--) {
 70         read(i), read(j), read(x);
 71         if (a[i] == 0) {
 72             puts("Yes"); continue;
 73         }
 74         if (x == 1) {
 75             puts("Yes"); continue;
 76         }
 77         if (a[i] == 1) {
 78             puts("No"); continue;
 79         }
 80         if (a[i+1] == 0 && j >= i + 1) {
 81             puts("No"); continue;
 82         }
 83         cnt = 0; flag = true;
 84         while (x != 1) {
 85             LL tmp = gcd(x, a[i]);
 86             if (tmp == 1) {
 87                 flag = false; break;
 88             }
 89             while (x % tmp == 0) x /= tmp, ++cnt;
 90         }
 91         if (near0[i] <= j) j = min(j, near0[i] - 2);
 92         if (near1[i] <= j) j = min(j, near1[i] - 1);
 93         if (j == i) {
 94             if (!flag || cnt > 1) puts("No");
 95             else if (a[i] % x == 0) puts("Yes");
 96             else puts("No");
 97             continue;
 98         }
 99         if (!flag || !ok(i+1, j)) puts("No");
100         else puts("Yes");
101     }
102     return 0;
103 }