Light oj 1085 - All Possible Increasing Subsequences (简单dp + 离散化 + BIT)

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1085

题意：

        问你有多少个上升子序列。

思路：

        dp[i]表示以第i个数结尾的上升序列数量。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <map>
 6 using namespace std;
 7 typedef long long LL;
 8 const int N = 1e5 + 5;
 9 int a[N];
10 int b[N], n;
11 LL bit[N], mod = 1e9 + 7;
12 LL dp[N];
13 
14 void add(int i, LL val) {
15     for( ; i <= n; i += (i&-i))
16         bit[i] += val;
17 }
18 
19 LL sum(int i) {
20     LL s = 0;
21     for( ; i >= 1; i -= (i&-i))
22         s += bit[i];
23     return s;
24 }
25 
26 int main()
27 {
28     int t;
29     scanf("%d", &t);
30     for(int ca = 1; ca <= t; ++ca) {
31         scanf("%d", &n);
32         for(int i = 1; i <= n; ++i) {
33             scanf("%d", a + i);
34             b[i] = a[i];
35         }
36         sort(b + 1, b + n + 1);
37         memset(bit, 0, sizeof(bit));
38         memset(dp, 0, sizeof(dp));
39         for(int i = 1; i <= n; ++i) {
40             a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;
41             dp[i]++;
42         }
43         for(int i = 1; i <= n; ++i) {
44             dp[i] += sum(a[i] - 1);
45             dp[i] %= mod;
46             add(a[i], dp[i]);
47         }
48         LL ans = 0;
49         for(int i = 1; i <= n; ++i) {
50             ans += dp[i];
51             ans %= mod;
52         }
53         printf("Case %d: %lld\n", ca, ans);
54     }
55     return 0;
56 }

 

posted @ 2016-10-09 17:18  Recoder  阅读(248)  评论(0)  编辑  收藏  举报