Hdu 5288 OO's Sequence (2015 first case in multiple schools, 1st questions) Enumeration factor, hdu5288

Hdu 5288 OO's Sequence (2015 first case in multiple schools, 1st questions) Enumeration factor, hdu5288

Question link:Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 5288

 

Question:In the closed interval [l, r], there is a number a [I], and a [I] cannot be the number except itself, f (l, r) it indicates the number of numbers in the range a [I]. Now we give you n numbers to calculate the sum of f (l, r) in all intervals.

 

Ideas:For each number a [I], locate the position L and R of the factor nearest to him on the left and right, and record it, then, each number a [I] has its L, R, and for each a [I] In f (l, r) the sum of values is (I-L + 1) * (R-I + 1)

 

Code:

 

1 # include <cstdio> 2 # include <cstdlib> 3 # include <cmath> 4 # include <cstring> 5 # include <iostream> 6 # include <queue> 7 # include <algorithm> 8 # include <vector> 9 using namespace std; 10 # define LL _ int6411 # define INF 0x3f3f3f3f12 const int MAXN = 100005; 13 # define mod 100000000714 int l [MAXN]; 15 int r [MAXN]; 16 int a [MAXN]; 17 int visl [MAXN]; 18 int visr [MAXN]; 19 LL ans; 20 int main () 21 {22 int n, I, J; 23 while (~ Scanf ("% d", & n) 24 {25 memset (visl, 0, sizeof (visl); 26 memset (visr, 0, sizeof (visr )); 27 for (I = 1; I <= n; I ++) 28 {29 scanf ("% d", & a [I]); 30 l [I] = 1; 31 r [I] = n; 32} 33 for (I = 1; I <= n; I ++) 34 {35 for (j = a [I]; j <= 10000; j + = a [I]) 36 {37 if (visr [j] & r [visr [j] = n) 38 r [visr [j] = I-1; 39} 40 visr [a [I] = I; 41} 42 for (I = n; I> 0; I --) 43 {44 for (j = a [I]; j <= 10000; j + = a [I]) 45 {46 if (visl [j] & l [visl [j] = 1) 47 l [visl [j] = I + 1; 48} 49 visl [a [I] = I; 50} 51 ans = 0; 52 for (I = 1; I <= n; I ++) 53 {54 // printf ("% d \ n", l [I], r [I]); 55 ans + = (LL) (I + 1-l [I]) * (r [I]-I + 1); 56 ans % = mod; 57} 58 printf ("% I64d \ n ", ans % mod); 59} 60 return 0; 61}View Code