# The Battle of Chibi

Time Limit: 6000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2899    Accepted Submission(s): 1043

Problem Description
Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loyal, it's impossible to convince any of them to betray Cao Cao.

So there is only one way left for Yu Zhou, send someone to fake surrender Cao Cao. Gai Huang was selected for this important mission. However, Cao Cao was not easy to believe others, so Gai Huang must leak some important information to Cao Cao before surrendering.

Yu Zhou discussed with Gai Huang and worked out

Input
The first line of the input gives the number of test cases,

Output
For each test case, output one line containing Case #x: y, where

Sample Input
2 3 2 1 2 3 3 2 3 2 1

Sample Output
Case #1: 3 Case #2: 0
Hint
In the first cases, Gai Huang need to leak 2 information out of 3. He could leak any 2 information as all the information value are in increasing order. In the second cases, Gai Huang has no choice as selecting any 2 information is not in increasing order.

Source

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题意：

问一个序列之中有多少个长度为M的严格递增子序列。

思路：

我们用dp[i][j]表示前j个数中，以Aj为结尾的长度为i的严格递增子序列的个数。

那么对于dp[i][j]，我们只需要枚举所有小于j的k，并且Ak < Aj，将所有的dp[i-1][k]求和，可以得到dp[i][j]

很容易想到O(n^3)的算法，但是显然会超时，所以我们需要进行一些优化。

当我们枚举内层循环j,k时，外层循环i就可以被当成是定值。当j增加1，k的取值只是多了k = j这个新的决策。

因此我们用树状数组维护一个前缀和，表示1~j区间，长度为i-1时的方案数。

最开始离散化用的set和map TLE了

后来改用了另一个数组，先排序然后lower_bound，并且直接把原数组的值改掉

 1 #include <iostream>
2 #include <set>
3 #include <cmath>
4 #include <stdio.h>
5 #include <cstring>
6 #include <algorithm>
7 #include <map>
8 using namespace std;
9 typedef long long LL;
10 #define inf 0x7f7f7f7f
11
12 int t, n, m, cnt;
13 const int maxn = 1005;
14 const LL mod = 1e9 + 7;
15 int a[maxn], b[maxn];
16 LL sum[maxn], dp[maxn][maxn];
17 map<LL, int>discrete;
18 set<LL>sss;
19 set<LL>::iterator iter;
20
21 void add(int pos, LL x)
22 {
23     while(pos <= n + 1){
24         sum[pos] = (sum[pos] + x) % mod;
25         pos += (pos & -pos);
26     }
27 }
28
30 {
31     LL ans = 0;
32     while(pos){
33         ans = (ans + sum[pos]) % mod;;
34         pos -= (pos & -pos);
35     }
36     return ans;
37 }
38
39 void init()
40 {
41     discrete.clear();
42     sss.clear();
43     //memset(dp, 0, sizeof(dp));
44     for(int i = 0; i <= m; i++){
45         for(int j = 0; j <= n; j++){
46             dp[i][j] = 0;
47         }
48     }
49     cnt = 0;
50 }
51
52 int main()
53 {
54     scanf("%d", &t);
55     for(int cas = 1; cas <= t; cas++){
56         init();
57         scanf("%d%d", &n, &m);
58         a = b = -inf;
59         dp = 1;
60         //sss.insert(a);
61         for(int i = 1; i <= n; i++){
62             scanf("%d", &a[i]);
63             b[i] = a[i];
64             //sss.insert(a[i]);
65         }
66         sort(b, b + n + 1);
67         for(int i = 0; i <= n; i++){
68             a[i] = lower_bound(b, b + 1 + n, a[i]) - b + 1;
69             //cout<<a[i]<<endl;
70         }
71         /*for(iter = sss.begin(); iter != sss.end(); iter++){
72             discrete[*iter] = ++cnt;
73         }*/
74         //cout<<a<<endl;
75         for(int i = 1; i <= m; i++){
76             //memset(sum, 0, sizeof(sum));
77             for(int j = 0; j <= n + 1; j++){
78                 sum[j] = 0;
79             }
81
82             for(int j = 1; j <= n; j++){
83                 dp[i][j] = ask(a[j] - 1);
84                 //if(discrete[a[j]] < discrete[a[j + 1]])
86             }
87         }
88
89         int ans = 0;
90         for(int i = 1; i <= n; i++){
91             ans = (ans + dp[m][i]) % mod;
92         }
93         printf("Case #%d: %d
", cas, ans);
94     }
95 }
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