树状数组 + dp

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 #include "algorithm"
 5 using namespace std;
 6 int n;
 7 struct st
 8 {
 9     int ans, rank, pos;
10 } a[100010];
11 int tot;
12 bool cmp_1(st a, st b)
13 {
14     return a.ans < b.ans;
15 }
16 bool cmp_2(st a, st b)
17 {
18     return a.pos < b.pos;
19 }
20 
21 inline int lowbit(int p)
22 {
23     return p&-p;
24 }
25 const int mod = 1000000007;
26 
27 const int maxn = 100010;
28 int t[maxn], element[maxn];
29 void add(int p, int x)
30 {
31     int i;
32     for(i = p; i <= tot; i += lowbit(i)) {
33         t[i] += x;
34         t[i] %= mod;
35     }
36 }
37 
38 __int64 sum(int p)
39 {
40     int i, res = 0;
41     for(i = p; i >= 1; i -= lowbit(i)) {
42         res += t[i];
43         res %= mod;
44     }
45     return res;
46 }
47 
48 int main()
49 {
50     int i, j;
51     scanf("%d", &n);
52     for(i = 1; i<=n; ++i) {
53         scanf("%d", &a[i].ans);
54         a[i].pos = i;
55     }
56     sort(a+1, a+1+n, cmp_1);
57     tot = 1;
58     a[1].rank = 1;
59     for(i = 2; i<=n; ++i) {
60         if(a[i].ans!=a[i-1].ans)
61             a[i].rank = ++tot;
62         else
63             a[i].rank = tot;
64     }
65     sort(a+1, a+1+n, cmp_2);
66     int tmp;
67     for(i = 1; i<=n; ++i) {
68         tmp = (sum(a[i].rank)*a[i].ans%mod+a[i].ans)%mod;   //a[i].ans 是对空串的延长，因为 树状数组 不能计算到 t[0]，不然就不用特殊处理空串的情况了
69         add(a[i].rank, (tmp-element[a[i].rank]+mod)%mod);   //减去重复计算值，即上一次的计算值
70         element[a[i].rank] = tmp;   //记录上一次的计算值
71     }
72     printf("%d\n", sum(tot));
73 }