add some template for ec-final

二维rmq 离线 init O( n*n*logn*logn )  query O(1)

http://www.cnblogs.com/kuangbin/p/3227420.html

求1-n有多少个素数 n = 1e9

 #include <bits/stdc++.h>
 using namespace std;
 namespace acm {
 /// count primes between [1..n] (0< n <=1e9)
 /// solution: dp
 /// d(n, i) means that the number of value whose min prime number is bigger than i in [1..n]
 /// d(n, i) = d(n, i-1) if number i is not a prime
 /// d(n, i) = d(n, i-1) - d(n/i, i) + d(i, i-1) if number i is a prime
 /// complex time: O(n^(3/4)), space: O(n^(1/2))
 struct CountPrime {
     int solve(int n) {
         if (n < ) {
             return ;
         }
         dp[] = ;
         int sqrtn = , length = ;
         for (; sqrtn <= n / sqrtn; ++sqrtn) {
             dp[++length] = sqrtn - ;
         }
         for (int i = sqrtn - ; i > ; --i) {
             int val = n / i;
             if (val != i) {
                 dp[++length] = val - ;
             }
         }
         for (int i = ; i < sqrtn; ++i) {
             if (dp[i] == dp[i-]) {
                 continue;
             }
             for (int j=length; j>; --j) {
                 int val = j < sqrtn ? j : n / (length - j + );
                 if (i > val / i) {
                     break;
                 }
                 int pre_statue_pos = val / i;
                 if (pre_statue_pos >= sqrtn) {
                     pre_statue_pos = length - n / pre_statue_pos + ;
                 }
                 dp[j] += dp[i-] - dp[pre_statue_pos];
             }
         }
         return dp[length];
     }
     static const int maxn = 1e5+;
     int dp[maxn];
 };
 } // namespace acm
 int main()
 {
     //freopen("dp.in", "r", stdin);
     //freopen("dp.out", "w", stdout);
     acm::CountPrime cp;
     int n;
     while (scanf ("%d", &n) != EOF) {
         printf ("%d\n", cp.solve(n));
     }
     return ;
 }

end