codeforces round #266 D Increase Sequence 记忆化搜索

最新推荐文章于 2021-09-09 22:41:11 发布

messyidea

最新推荐文章于 2021-09-09 22:41:11 发布

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Peter has a sequence of integers a1, a2, ..., an. Peter wants all numbers in the sequence to equal h. He can perform the operation of "adding one on the segment [l, r]": add one to all elements of the sequence with indices from l to r (inclusive). At that, Peter never chooses any element as the beginning of the segment twice. Similarly, Peter never chooses any element as the end of the segment twice. In other words, for any two segments [l1, r1] and [l2, r2], where Peter added one, the following inequalities hold: l1 ≠ l2 and r1 ≠ r2.

How many distinct ways are there to make all numbers in the sequence equal h? Print this number of ways modulo 1000000007 (109 + 7). Two ways are considered distinct if one of them has a segment that isn't in the other way.

Input

The first line contains two integers n, h (1 ≤ n, h ≤ 2000). The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 2000).

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

        Sample test(s) 
      
          input 
        
3 2
1 1 1

          output 
        
4

          input 
        
5 1
1 1 1 1 1

          output 
        
1

          input 
        
4 3
3 2 1 1

          output 
        
0

有n个数，让它们全部到达m,每次可以选一段，增加1.求种类数。每次选的段左右坐标不能相同。

记忆化搜索

类似于加括号。

每次左边有多少k未匹配的线段，如果它本身加上这些线段可以等于m,那么就匹配一个，然后后面还剩下k个未匹配。

也可能它再加上一条线段才能到达m,那么在它上面再加一个为匹配的线段。

#include <cstdlib>
#include <climits>
#include <cassert>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <functional>
#include <algorithm>
#include <memory.h>
#include <numeric>
#include <utility>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <fstream>

#include <cstring>
#include <vector>
#include <bitset>
#include <deque>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#define MOD 1000000007
#define MX 2005
using namespace std;

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

LL val[MX][MX];
int N, H, A[MX];

LL calc(int n, int k) {     //左边有k个未匹配(先匹配和后匹配没关系)
    LL &ret = val[n][k];
    if (ret != -1) return ret;
    if (n == N) {
        if (k == 0) return ret = 1;
        return ret = 0;
    }
    ret = 0;
    if (A[n] + k == H) {
        if (k) ret = k * calc(n + 1, k - 1);
        ret = (ret + calc(n + 1, k)) % MOD;
        return ret;
    }
    if (A[n] + k + 1 == H) return ret = (calc(n + 1, k) * (k + 1) + calc(n + 1, k + 1)) % MOD;
    return ret;
}

main() {
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    
    memset(val, -1, sizeof val);
    scanf("%d%d", &N, &H);
    for (int i = 0; i < N; i++) scanf("%d", A + i);
    printf("%d\n", (int)calc(0, 0));
}