hdoj 5568 sequence2 【dp + 高精度】

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笑着走完自己的路

最新推荐文章于 2022-03-03 19:35:46 发布

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分类专栏：我就是DP 高精度

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本文深入探讨了如何通过动态规划解决计算给定整数数组中长度为k的严格递增子序列数量的问题。详细介绍了算法的核心思想，包括状态转移方程的设定和高精度计算的实现，最终通过样例输入输出展示了算法的实际应用。

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sequence2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 124 Accepted Submission(s): 55

Problem Description

Given an integer array

bi with a length of

n, please tell me how many exactly different increasing subsequences.

P.S. A subsequence

bai(1≤i≤k) is an increasing subsequence of sequence

bi(1≤i≤n) if and only if

1≤a1<a2<...<ak≤n and

ba1<ba2<...<bak.
Two sequences

ai and

bi is exactly different if and only if there exist at least one

i and

ai≠bi.

Input

Several test cases(about

5)

For each cases, first come 2 integers,

n,k(1≤n≤100,1≤k≤n)

Then follows

n integers

ai(0≤ai≤109)

Output

For each cases, please output an integer in a line as the answer.

Sample Input

Sample Output

2
3

题意：给定n个元素组成的序列a[]，问你长度为k的严格递增子序列的个数。

很水的dp，只不过需要高精度。。。模版出了点问题，手写了高精度。

思路：定义dp[i][j]为以a[i]结尾的长度为j胡严格递增子序列个数。

dp[i][j] += sigma(dp[p][j-1]) 其中(1 <= p < i && a[p] < a[i])。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define MAXN 100
#define MAXM 1010
#define eps 1e-8
#define LL long long
#define MAXSIZE 1005
using namespace std;
struct bign
{
    int s[MAXSIZE];
    void operator += (const bign &t)
    {
        for(int i = 0; i < MAXSIZE; i++)
        {
            s[i] += t.s[i];
            if(s[i] >= 10)
            {
                s[i] -= 10;
                s[i+1]++;
            }
        }
    }
    void put()
    {
        int i;
        for(i = MAXSIZE - 1; i >= 0; i--)
            if(s[i])
                break;
        if(i == -1)
            printf("0");
        else
            for(; i >= 0; i--)
                printf("%d", s[i]);
    }
};
bign dp[101][101];
int a[101];
int main()
{
    int n, k;
    while(scanf("%d%d", &n, &k) != EOF)
    {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= k; j++)
                memset(dp[i][j].s, 0, sizeof(dp[i][j].s));
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]), dp[i][1].s[0] = 1;
        for(int i = 1; i <= n; i++)
            for(int j = 2; j <= min(i, k); j++)
                for(int p = 1; p < i; p++)
                    if(a[p] < a[i])
                        dp[i][j] += dp[p][j-1];
        bign ans; memset(ans.s, 0, sizeof(ans.s));
        for(int i = 1; i <= n; i++)
            ans += dp[i][k];
        ans.put();
        printf("\n");
    }
    return 0;
}