[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

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <cmath>
 7 #include <queue>
 8 #include <map>
 9 #include <stack>
10 #include <list>
11 #include <vector>
12
13 using namespace std;
14
15 inline int max(int a, int b) {
16     return a > b ? a : b;
17 }
18
19 const int maxn = 100010;
20 const int mod = 1000000007;
21 int a[maxn];
22 int s[maxn];
23 int l[maxn];
24 int r[maxn];
25 int n;
26 long long ans;
27
28 void left() {
29     int m = 0;
30     memset(s, 0, sizeof(s));
31     for(int i = n; i >= 1 ; i--) {
32         m = max(m, a[i]);
33         for(int j = a[i]; j <= m; j+=a[i]) {
34             if(s[j] && l[s[j]] < i) {
35                 l[s[j]] = i;
36             }
37             s[a[i]] = i;
38         }
39     }
40 }
41
42 void right() {
43     int m = 0;
44     memset(s, 0, sizeof(s));
45     for(int i = 1; i <= n; i++) {
46         m = max(m, a[i]);
47         for(int j = a[i]; j <= m; j+=a[i]) {
48             if(s[j] && r[s[j]] > i) {
49                 r[s[j]] = i;
50             }
51             s[a[i]] = i;
52         }
53     }
54 }
55 int main() {
56     freopen("in", "r", stdin);
57     while(~scanf("%d", &n)) {
58         ans = 0;
59         for(int i = 1; i <= n; i++) {
60             scanf("%d", &a[i]);
61             l[i] = 0;
62             r[i] = n + 1;
63         }
64         right();
65         left();
66         // for(int i = 1; i <= n; i++) {
67         //     printf("%d %d %d\n", l[i], a[i], r[i]);
68         // }
69         for(int i = 1; i <= n; i++) {
70             ans = (ans + ((i - l[i]) % mod * (r[i] - i)) % mod) % mod;
71         }
72         printf("%I64d\n", ans);
73     }
74 }

[HDOJ5288]OO's Sequence