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 }
Hackerrank--Divisibility of Power(Math),布布扣,bubuko.com
Hackerrank--Divisibility of Power(Math)
原文：http://www.cnblogs.com/Stomach-ache/p/3917691.html