Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)
(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)
Since the answer may be large, return the answer modulo 10^9 + 7.
Example 1:
Input: n = 5 Output: 12 Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.
Example 2:
Input: n = 100 Output: 682289015
Constraints:
1 <= n <= 100
Related Topics:
Math
// OJ: https://leetcode.com/problems/prime-arrangements/
// Author: github.com/lzl124631x
// Time: O(NloglogN)
// Space: O(N)
class Solution {
int countPrimes(int n) {
if (n <= 1) return 0;
vector<int> prime(n + 1, true);
int ans = 0, bound = sqrt(n);
for (int i = 2; i <= n; ++i) {
if (!prime[i]) continue;
++ans;
if (i > bound) continue;
for (int j = i * i; j <= n; j += i) prime[j] = false;
}
return ans;
}
public:
int numPrimeArrangements(int n) {
int cnt = countPrimes(n), ans = 1, mod = 1e9 + 7;
for (int i = 1; i <= cnt; ++i) ans = ((long)ans * i) % mod;
for (int i = 1; i <= n - cnt; ++i) ans = ((long)ans * i) % mod;
return ans;
}
};