[HDOJ5288]OO's Sequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5288

转换成求从1到n可以整除A[i]的区间的数量。
找离最近的A[i]最近的可以整除A[i]的l和r，因为再往左或者右必然包含l和r
假如[l,i]有x个数字，[i,r]有y个数字，那么A[i]的贡献即(x+1)*(y+1)
求A[i]对ans的贡献综合
即求离A[i]最近的可以被A[i]整除的数

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>

using namespace std;

inline int max(int a, int b) {
    return a > b ? a : b;
}

const int maxn = 100010;
const int mod = 1000000007;
int a[maxn];
int s[maxn];
int l[maxn];
int r[maxn];
int n;
long long ans;

void left() {
    int m = 0;
    memset(s, 0, sizeof(s));
    for(int i = n; i >= 1 ; i--) {
        m = max(m, a[i]);
        for(int j = a[i]; j <= m; j+=a[i]) {
            if(s[j] && l[s[j]] < i) {
                l[s[j]] = i;
            }
            s[a[i]] = i;
        }
    }
}

void right() {
    int m = 0;
    memset(s, 0, sizeof(s));
    for(int i = 1; i <= n; i++) {
        m = max(m, a[i]);
        for(int j = a[i]; j <= m; j+=a[i]) {
            if(s[j] && r[s[j]] > i) {
                r[s[j]] = i;
            }
            s[a[i]] = i;
        }
    }
}
int main() {
    freopen("in", "r", stdin);
    while(~scanf("%d", &n)) {
        ans = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            l[i] = 0;
            r[i] = n + 1;
        }
        right();
        left();
        // for(int i = 1; i <= n; i++) {
        //     printf("%d %d %d
", l[i], a[i], r[i]);
        // }
        for(int i = 1; i <= n; i++) {
            ans = (ans + ((i - l[i]) % mod * (r[i] - i)) % mod) % mod;
        }
        printf("%I64d
", ans);
    }
}