Hackerrank--Divisibility of Power(Math)

题目链接

You are given an array A of size N. You are asked to answer Q queries.

Each query is of the form :

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≤2×105 
2≤Q≤3×105 
1≤i,j≤N 
i≤j 
1≤x≤1016 
0≤ value of array element ≤1016

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:

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 using namespace std;
 const int maxn = ;
 typedef long long LL;
 const int inf = 0x3f3f3f3f;
 LL a[maxn], x;
 int n, Q, i, j, cnt, near0[maxn], near1[maxn], c[maxn];
 bool flag;
 
 void init() {
     memset(near0, 0x3f, sizeof(near0));
     memset(near1, 0x3f, sizeof(near1));
     cnt = ;
     for (i = ; i <= n; i++) if (a[i] == ) c[cnt++] = i;
     int be = ;
     for (i = ; i < cnt; i++) {
         for (j = be; j < c[i]; j++) near0[j] = c[i];
         be = c[i] + ;
     }
     cnt = ;
     for (i = ; i <= n; i++) if (a[i] == ) c[cnt++] = i;
     be = ;
     for (i = ; i < cnt; i++) {
         for (j = be; j < c[i]; j++) near1[j] = c[i];
         be = c[i] + ;
     }
 }
 
 void read(LL &res) {
     res = ;
     char c = ' ';
     while (c < '' || c > '') c = getchar();
     while (c >= '' && c <= '') res = res *  + c - '', c = getchar();
 }
 void read(int &res) {
     res = ;
     char c = ' ';
     while (c < '' || c > '') c = getchar();
     while (c >= '' && c <= '') res = res *  + c - '', c = getchar();
 }
 
 LL gcd(LL a, LL b) {
     if (!b) return a;
     else return gcd(b, a % b);
 }
 
 bool ok(int be, int en) {
     LL res = ;
     for (int i = en; i >= be; i--) {
         //if (quick_pow(a[i], res, res)) return true;
         LL tmp = ;
         for (int j = ; j < res; j++) {
             if (tmp >= cnt || a[i] >= cnt) return true;
             tmp *= a[i];
         }
         if (tmp >= cnt) return true;
         res = tmp;
     }
     return res >= cnt;
 }
 int main(void) {
     read(n);
     for (i = ; i <= n; i++) read(a[i]);
     init();
     read(Q);
     while (Q--) {
         read(i), read(j), read(x);
         if (a[i] == ) {
             puts("Yes"); continue;
         }
         if (x == ) {
             puts("Yes"); continue;
         }
         if (a[i] == ) {
             puts("No"); continue;
         }
         if (a[i+] ==  && j >= i + ) {
             puts("No"); continue;
         }
         cnt = ; flag = true;
         while (x != ) {
             LL tmp = gcd(x, a[i]);
             if (tmp == ) {
                 flag = false; break;
             }
             while (x % tmp == ) x /= tmp, ++cnt;
         }
         if (near0[i] <= j) j = min(j, near0[i] - );
         if (near1[i] <= j) j = min(j, near1[i] - );
         if (j == i) {
             if (!flag || cnt > ) puts("No");
             else if (a[i] % x == ) puts("Yes");
             else puts("No");
             continue;
         }
         if (!flag || !ok(i+, j)) puts("No");
         else puts("Yes");
     }
     return ;
 }