[Explanations] Luo Gu P4821 [Zhongshan election] Spanning Tree

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Subject description:
There is a graphic called a pentagonal ring.

A circle with a center pentagon an n by n vertices and edges consisting ring. At the same time also a certain side of a pentagon, a total of n different in each of the n pentagonal rim midpoint of the side. The pentagon only common vertex on the center of the ring pentagon circle. FIG pentagonal ring is a 4- 0 FIG.
N now given a pentagonal ring, your task is the number of different spanning n pentagonal rings obtained. Remember what is the spanning tree do? It is a spanning tree diagram and the number of all the vertexes of the original reservation subtracting a plurality of such edge, thereby generating a tree.
Note: n pentagon in a given circle all the vertices are as different vertices.

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Input format
input comprises a plurality of sets of test data.
The first line contains a positive integer T, represents the number of test data.
Each test contains an integer n (2 <= N <= 100), you need to solve the edge represents the number of turns of the center pentagon.

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Output format
for each set of test data output line contains an integer x, denotes the number of spanning trees results after 2007 pentagonal ring modulo n.

Analysis:
a mathematical derivation problem ( $wc$ is actually placed in graph theory, specifically about. . ).

• We first consider the concept of spanning tree: $n$ FIG points are $n-1$ edges are attached.

Not difficult to draw, drawing a total of $4n$ points, $5n$ sides.

• Depending on the nature of the spanning tree, we need to remove $n-1$ edges.

Shared equally down each side of a pentagon need to be deleted, leaving an edge. (That there is a need to delete the two sides of the pentagon)

Since the positive FIG. $n$ deformation is a ring, so we must have a positive side is the need to delete $n$ -sided polygon edges.

• A pentagon need to delete the two sides, there is a positive side to be deleted $n$ -gon side, there are four ways, each pentagon may be deleted two sides of this pentagon, a total of $n$ a is $4n$ ways
• The remaining n-1 each have five sides pentagons can be deleted, a total of $5^{n-1}$ kinds of programs

Therefore, the total number of programs are:
$4*n\ +\ 5^{n-1}$

Then the specific code as follows:

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define P 2007
int t;
int main(){
	freopen("data.in","r",stdin);
	freopen("data.out","w",stdout);
	scanf("%d",&t);
	while (t--){
	    int n;
	    scanf("%d",&n);
	    LL Pow=1;
	    for (int i=1;i<n;i++) Pow=(Pow*5)%P;
	    printf("%lld\n",(Pow*4*n)%P);
	}
	fclose(stdin);
	fclose(stdout);
	return 0;
}