2015 Multi-University Training Contest 1 - 1001 OO’s Sequence

 OO’s Sequence#

Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5288
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Mean: 

给定一个数列，让你求所有区间上满足Ai%Aj!=0（Ai!=Aj）的Ai的个数之和。

analyse:

对于Ai，如果我们知道最靠近Ai且能够整除Ai的数的下标l和r，那么Ai对答案的贡献就是(r-i)*(i-l)。剩下的就是怎样去求每个Ai的l和r了。

首先我们预处理出：对于每个i，能够被1~i整除的数，用链表存起来。

那么对于输入的数列Ai，我们就可以在O(1)的时间复杂度内知道他能够被哪些数整除，然后去找这些数在pos数组中映射的位置。

从左往右求出每个Ai的l，从右往左求出每个Ai的r，然后O(n)扫一遍统计答案。

Time complexity: O(N*sqrt(A))

 

Source code: 

 

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/* * this code is made by crazyacking * Verdict: Accepted * Submission Date: 2015-07-22-08.50 * Time: 0MS * Memory: 137KB */ #include <queue> #include <cstdio> #include <set> #include <string> #include <stack> #include <cmath> #include <climits> #include <map> #include <cstdlib> #include <iostream> #include <vector> #include <algorithm> #include <cstring> #define  LL long long #define  ULL unsigned long long using namespace std; const int MAXN = 10005, MAXX = 100005, mod = 1e9 + 7; vector<int> divi[MAXN]; int a[MAXX], l[MAXX], r[MAXX], pos[MAXX], n;   void init() {       for( int i = 1; i <= 10000; ++i )             for( int j = 1; j <= i; ++j )                   if( !( i % j ) ) divi[i].push_back( j ); }   int main() {       ios_base::sync_with_stdio( false );       cin.tie( 0 );       init();       while( cin >> n )       {             for( int i = 0; i < n; ++i ) cin >> a[i];             memset( l, -1, sizeof l );             memset( r, 0x3f, sizeof r );             memset( pos, -1, sizeof pos );             for( int i = 0; i < n; ++i )             {                   int lef = -1;                   for( int j = 0; j < divi[a[i]].size(); ++j )                         lef = max( lef, pos[divi[a[i]][j]] );                   pos[a[i]] = i;                   l[i] = lef;             }             memset( pos, 0x3f, sizeof pos );             for( int i = n - 1; i >= 0; --i )             {                   int rig = 0x3f3f3f3f;                   for( int j = 0; j < divi[a[i]].size(); ++j )                         rig = min( rig, pos[divi[a[i]][j]] );                   pos[a[i]] = i;                   r[i] = rig;             }             int ans = 0, L, R;             for( int i = 0; i < n; ++i )             {                   if( l[i] == -1 ) L = i + 1;                   else L = i - l[i];                   if( r[i] == 0x3f3f3f3f ) R = n - i;                   else R = r[i] - i;                   ans = ( L * R % mod + ans ) % mod;             }             cout << ans << endl;       }       return 0; } /*   */