hdu 5288 (预处理+暴力) OO’s Sequence

题意：

见题面。

思路

预处理出每个a[i]左边和右边第一个能整除它的位置L[i]和R[i],然后计算这个值对于答案的贡献的个数。
贡献就是左右区间长度相乘

参考code:

/*
 #pragma warning (disable: 4786)
 #pragma comment (linker, "/STACK:0x800000")
 */
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <iterator>
#include <utility>
using namespace std;

template< class T > T _abs(T n) { return (n < 0 ? -n : n); }
template< class T > T _max(T a, T b) { return (!(a < b) ? a : b); }
template< class T > T _min(T a, T b) { return (a < b ? a : b); }
template< class T > T sq(T x) { return x * x; }
template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }
template< class T > bool inside(T a, T b, T c) { return a<=b && b<=c; }


#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define F(i, n) for(int (i)=0;(i)<(n);++(i))
#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))
#define repok(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))
#define MEM0(addr) memset((addr), 0, sizeof((addr)))
#define MP(x, y) make_pair(x, y)
#define REV(s, e) reverse(s, e)
#define SET(p) memset(pair, -1, sizeof(p))
#define CLR(p) memset(p, 0, sizeof(p))
#define MEM(p, v) memset(p, v, sizeof(p))
#define CPY(d, s) memcpy(d, s, sizeof(s))
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
#define SZ(c) (int)c.size()
#define PB(x) push_back(x)
#define ff first
#define ss second
#define ll long long
#define ld long double
#define pii pair< int, int >
#define psi pair< string, int >
#define ls u << 1
#define rs u << 1 | 1
#define lson l, mid, u << 1
#define rson mid, r, u << 1 | 1
#define debug(x) cout << #x << " = " << x << endl

const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int MAXN = 1e5+5;
const int mod = 1e9 + 7;

int R[MAXN],L[MAXN];
int a[MAXN];
std::vector<int> pos[10010];
int n;
void init(){
    rep(i,0,10005) pos[i].clear();
    rep(i,0,n-1){L[i] = 0; R[i] = n - 1;}
}
void solve(){
    rep(i,0,n-1){
        for(int j = a[i]; j <= 10000 ; j += a[i]){
            int len = pos[j].size();
            rep(k,0,len-1){
                int x = pos[j][k];
                if(x == i) continue;
                else if(x > i) L[x] = max(L[x],i+1);
                else R[x] = min(R[x],i-1);
            }
        }
    }
}
int main(){
    //READ("in.txt");
    while(scanf("%d",&n)!=EOF){
        init();
        rep(i,0,n-1){
            scanf("%d",&a[i]);
            pos[a[i]].push_back(i);
        }
        solve();
        ll ans = 0;
        rep(i,0,n-1){
            //debug(L[i]);
            //debug(R[i]);
            ll l = i - L[i] + 1LL;
            ll r = R[i] - i + 1LL;
            ans = (ans + (l * r) % mod) % mod;
        } 
        cout << ans << endl;
    }
}