链接:

https://codeforces.com/gym/102394/problem/I

题意：

DreamGrid has an interesting permutation of 1,2,…,n denoted by a1,a2,…,an. He generates three sequences f, g and h, all of length n, according to the permutation a in the way described below:

For each 1≤i≤n, fi=max{a1,a2,…,ai};
For each 1≤i≤n, gi=min{a1,a2,…,ai};
For each 1≤i≤n, hi=fi−gi.
BaoBao has just found the sequence h DreamGrid generates and decides to restore the original permutation. Given the sequence h, please help BaoBao calculate the number of different permutations that can generate the sequence h. As the answer may be quite large, print the answer modulo 109+7.

思路：

考虑，某个位置为n或者h[i] > h[i+1] ，和第一个位置不为0时，答案都为0.
其他情况，h[i] == h[i+1],考虑前面的空位数， h[i] < h[i+1],可以插最小值，也可以插最大值， 直接×2（没太懂）。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
 
const int MAXN = 1e5+10;
const int MOD = 1e9+7;
 
LL dp[MAXN];
int a[MAXN];
int n;
 
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d", &n);
        a[0] = 0;
        bool flag = true;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d", &a[i]);
            if (a[i] < a[i-1] || a[i] >= n || (i != 1 && a[i] == 0))
                flag = false;
        }
        if (a[1] != 0)
            flag = false;
        if (!flag)
        {
            puts("0");
            continue;
        }
        dp[1] = 1;
        for (int i = 2;i <= n;i++)
        {
            if (a[i] > a[i-1])
                dp[i] = (dp[i-1]*2)%MOD;
            else
            {
                LL sp = a[i]-i+2;
                dp[i] = (dp[i-1]*sp)%MOD;
            }
        }
        printf("%d\n", (int)(dp[n]%MOD));
    }
 
    return 0;
}