Clarke and tree

题目连接：

Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a CS, did a research on data structure.

Now Clarke has n nodes, he knows the degree of each node no more than ai. He wants to know the number of ways to choose some nodes to compose to a tree of size s(1≤s≤n).

Input

The first line contains one integer T(1≤T≤10), the number of test cases.

For each test case:

The first line contains an integer n(2≤n≤50).

Then a new line follow with n numbers. The ith number ai(1≤ai<n) denotes the number that the degree of the ith node must no more than ai.

Output

For each test case, print a line with n integers. The ith number denotes the number of trees of size i modulo 109+7.

Sample Input

1

3

2 2 1

Sample Output

3 3 2

Hint:

At first we know the degree of node 1 can not more than 2, node 2 can not more than 2, node 3 can not more than 1. So

For the trees of size 1, we have tree ways to compose, are 1, 2 and 3. i.e. a tree with one node.

For the trees of size 2, we have tree ways to compose, are 1-2, 1-3, 2-3.

For the trees of size 3, we have two ways to compose, are 1-2-3, 2-1-3.

题意

给你n个点，每个点的度数最多为a[i]

然后分别问你点数为s的树，一共有多少种，(1<=s<=n)

题解：

考虑prufer的序列

对于点数为s，在这道题中z可以转化为这样：长度为s-2的序列里面，s个数都出现小于a[i]次的序列个数有多少个

我们直接n^4dp就好了

dp[i][j][k]表示考虑到了第i个数，我用了j，当前序列的长度为k的方案数

代码

#include
    
     using namespace std;const int mod = 1e9+7;const int maxn = 62;long long dp[maxn][maxn][maxn];//考虑到了第i个点，现在出现了j个数，长度为k的方案数long long c[maxn][maxn];int a[maxn],n;void init(){    for(int i=0;i