hdu 5288 思路题

如果枚举区间的话是O(n^2)的复杂度一定会T，所以考虑每个数a[i]对答案的贡献，则只需找到a[i]的约数中在左边最靠近a[i]的和在右边最靠近a[i]的那两个约数位置，然后扫一遍统计答案即可。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

typedef long long ll;
const int INF = 0x3f;
const int MOD = 1000000007;
const int N = 100001;
int a[N];
int l[N];
int r[N];
int mp[N];

int main ()
{
    int n;
    while ( scanf("%d", &n) != EOF )
    {
        for ( int i = 1; i <= n; i++ )
        {
            scanf("%d", a + i);
            l[i] = 0;
            r[i] = n + 1;
        }
        memset( mp, -1, sizeof(mp) );
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j * j <= a[i]; j++ )
            {
                if ( a[i] % j == 0 )
                {
                    l[i] = max( l[i], mp[j] );
                    l[i] = max( l[i], mp[a[i] / j] );
                }
            }
            mp[a[i]] = i;
        }
        memset( mp, INF, sizeof(mp) );
        for ( int i = n; i >= 1; i-- )
        {
            for ( int j = 1; j * j <= a[i]; j++ )
            {
                if ( a[i] % j == 0 )
                {
                    r[i] = min( r[i], mp[j] );
                    r[i] = min( r[i], mp[a[i] / j] );
                }
            }
            mp[a[i]] = i;
        }
        int ans = 0;
        for ( int i = 1; i <= n; i++ )
        {
            ans = ( ans + ( ll ) ( i - l[i] ) * ( r[i] - i ) ) % MOD;
        }
        printf("%d
", ans);
    }
    return 0;
}