Hackerrank--Divisibility of Power(Math)
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
Yesif x divides the value returned from find(i,j) function, otherwise printNo.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
YesorNoas 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 ≤1016No 2 consecutive entries in the array will be zero.
Sample Input
4 2 3 4 5 2 1 2 4 1 3 7Sample 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 }