Codeforces Round #266 (Div. 2) D

D. Increase Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

复制题目占字数

真是NICE的题目。很考验思路

我的参考博客来一枚：http://blog.csdn.net/rowanhaoa/article/details/39343525

我再解释一下：先预处理一下,a[i]=h-a[i];

然后定义 b[i]=a[i]-a[i-1];(1<=i<=n+1）；

假如b[i]的值 -1<=b[i]<=1,如果有数不满足答案为0；

为什么，举个例子 n=4

a:  1 2 3 2;

处理后   b: 1 1 1 -1 -2;最后一项为-2，所以答案为0；

其实我们可以发现：最后a[n]一定<=1 因为它只能做一次在末尾，在中间的情况类推。

基本就是前面博主的正文了

 #include<stdio.h>
 #include<string.h>
 #include<iostream>
 using namespace std;
 typedef long long ll;
 #define mod 1000000007
 #define N 2222
 int a[N],b[N];
 
 int main()
 {
     int n,h;
     cin>>n>>h;
     ll ans=;
     ll cnt=;
     for (int i=;i<=n;i++) cin>>a[i],a[i]=h-a[i];
     for (int i=;i<=n+;i++) b[i]=a[i]-a[i-];
 
     for (int i=;i<=n+;i++)
     {
         if (b[i]==) cnt++;
         else if (b[i]==) ans=ans*(cnt+)%mod;
         else if (b[i]==-) ans=ans*cnt%mod,cnt--;
         else
         {
             ans=;
             break;
         }
     }
     cout<<ans<<endl;
     return ;
 }