The Battle of Chibi

Time Limit: 6000/4000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1905 Accepted Submission(s): 669

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

N information to be leaked, in happening order. Each of the information was estimated to has

ai value in Cao Cao's opinion.

Actually, if you leak information with strict increasing value could accelerate making Cao Cao believe you. So Gai Huang decided to leak exact

M information with strict increasing value in happening order. In other words, Gai Huang will not change the order of the

N information and just select

M of them. Find out how many ways Gai Huang could do this.

Input

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

T(1≤100).

T test cases follow.

Each test case begins with two numbers

N(1≤N≤103) and

M(1≤M≤N), indicating the number of information and number of information Gai Huang will select. Then

N numbers in a line, the

ith number

ai(1≤ai≤109) indicates the value in Cao Cao's opinion of the

ith information in happening order.

Output

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

x is the test case number (starting from 1) and

y is the ways Gai Huang can select the information.

The result is too large, and you need to output the result mod by

1000000007(109+7).

Sample Input

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

The 2015 China Collegiate Programming Contest

解題思路：dp[i][j]表示考慮到第i個數，且以第a[i]個數結尾，長度爲j的遞增序列個數，顯然dp[i][j] = sum(dp[k][j - 1]) a[i] > a[k],直接求解需要O(n * n * n)，顯然不可行，我們把數值離散化一下，然後用樹狀數組維護前綴和就行。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000 + 10;
typedef long long LL;
const LL mod = 1e9 + 7;
int N, M;
int a[maxn];
LL Tree[maxn][maxn];
LL dp[maxn][maxn];//dp[i][j]表示考慮到第i個數，且以第a[i]個數結尾，長度爲j的遞增序列個數
struct node{
    int value;
    int id;
    bool operator <(const node &res) const{
        if(value == res.value) return id > res.id;
        else return value < res.value;
    }
}Node[maxn];
int Rank[maxn];
void init()
{
    memset(Tree, 0, sizeof(Tree));
    memset(dp, 0, sizeof(dp));
}
int lowbit(int x)
{
    return x&(-x);
}
void add(int loc, int d, LL value)
{
    for(int i = loc; i <= N; i += lowbit(i))
    {
        Tree[i][d] = (Tree[i][d] + value) % mod;
    }
}
LL get(int loc, int d)
{
    LL ans = 0;
    for(int i = loc; i >= 1; i -= lowbit(i))
    {
        ans = (ans + Tree[i][d]) % mod;
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d", &T);
    int Case = 1;
    while(T--)
    {
        scanf("%d%d", &N, &M);
        init();
        for(int i = 1; i <= N; i++)
        {
            scanf("%d", &Node[i].value);
            Node[i].id = i;
        }
        sort(Node + 1, Node + N + 1);
        for(int i = 1; i <= N; i++)
        {
            Rank[Node[i].id] = i;
        }
        for(int i = 1; i <= N; i++)
        {
            dp[i][1] = 1;
            add(Rank[i], 1, 1);
            for(int j = 2; j <= min(M, i); j++)
            {
                LL temp = get(Rank[i] - 1, j - 1);
                dp[i][j] = (dp[i][j] + temp) % mod;
                add(Rank[i], j, dp[i][j]);
            }
        }
        LL ans = 0;
        for(int i = 1; i <= N; i++)
        {
            ans = (ans + dp[i][M]) % mod;
        }
        printf("Case #%d: %lld\n", Case++, ans);
    }
    return 0;
}