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:

 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 }