HDU 2372 ||SDUT 2384 El Dorado（动态规划）

El Dorado

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 304    Accepted Submission(s): 153

Problem Description

Bruce Force has gone to Las Vegas, the El Dorado for gamblers. He is interested especially in one betting game, where a machine forms a sequence of n numbers by drawing random numbers. Each player should estimate beforehand, how many increasing subsequences of length k will exist in the sequence of numbers. 

Bruce doesn't trust the Casino to count the number of increasing subsequences of length k correctly. He has asked you if you can solve this problem for him.

 

Input

The input contains several test cases. The first line of each test case contains two numbers n and k (1 ≤ k ≤ n ≤ 100), where n is the length of the sequence drawn by the machine, and k is the desired length of the increasing subsequences. The following line contains n pairwise distinct integers a_i (-10000 ≤ a_i ≤ 10000 ), where a_iis the i^th number in the sequence drawn by the machine. 

The last test case is followed by a line containing two zeros. 

Output

For each test case, print one line with the number of increasing subsequences of length k that the input sequence contains. You may assume that the inputs are chosen in such a way that this number fits into a 64 bit signed integer . 

Sample Input

10 5

1 2 3 4 5 6 7 8 9 10

3 2

3 2 1

0 0

Sample Output

252

0

Source

2008水题公开赛（比速度，OJ压力测试）

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lcy

 解题报告：这道题是一道动态规划题，题意就是求递增长度为m的总的个数，状态方程为dp[i][j] += dp[k][j - 1];其中dp[i][j]的含义：到第i个数时递增长度为j的个数;

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 110;
__int64 dp[N][N], ans;//dp[i][j]表示到第i个数的时候递增个数为j的个数
int m, n, a[N];
void DP()
{
    int i, j, k;
    memset(dp, 0, sizeof(dp));
    for (i = 1; i <= n; ++i)
    {
        dp[i][1] = 1;
    }
    for (j = 2; j <= m; ++j)//找长度为j的增长序列
    {
        for (i = j; i <= n; ++i)//遍历一次
        {
            for (k = j -1; k < i; ++k)//比较时应该是后面的和前面的作比较k<i作为条件
            {
                if (a[i] > a[k])
                {
                    dp[i][j] += dp[k][j - 1];//状态方程dp[i][j]表示第i数时，递增长度为j的个数
                }
            }
        }
    }
}
int main()
{
    int  i;
    while (scanf("%d%d", &n, &m) != EOF && n && m)
    {
        memset(a, 0, sizeof(a));
        for (i = 1; i <= n; ++i)
        {
            scanf("%d", &a[i]);
        }
        DP();
        ans = 0;
        for(i = m; i <= n; ++i)
        {
            ans += dp[i][m];//把递增长度为m的序列数全加起来
        }
        printf("%I64d\n", ans);
    }
    return 0;
}