Lily: “Chantarelle was part of my exotic phase.”

Buffy: “It’s nice. It’s a mushroom.”

Lily: “It is? That’s really embarrassing.”

Buffy: “Well, it’s an exotic mushroom, if that’s any comfort.”

Joss Whedon, "Anne".

A little girl whose name is Anne Spetring likes to play the following game. She draws a circle on

paper. Then she draws another one and connects it to the first cicrle by a line. Then she draws another

and connects it to one of the first two circles by a line. She continues this way until she has n circles

drawn and each one connected to one of the previously drawn circles. Her circles never intersect and

lines never cross. Finally, she numbers the circles from 1 to n in some random order.

How many different pictures can she draw that contain exactly n circles? Two pictures are different

if one of them has a line connecting circle number i to circle number j, and the other picture does not.

Input

The first line of input gives the number of cases, N. N test cases follow. Each one is a line containing

n (0 < n ≤ 100).

Output

For each test case, output one line containing ‘Case #x:’ followed by X, where X is the remainder

after dividing the answer by 2000000011.

Sample Input

3

1

2

3

Sample Output

Case #1: 1

Case #2: 1

Case #3: 3

?

題意：題目說畫圈，實(shí)際上圈可以看成是點(diǎn)，然后就是求n個(gè)點(diǎn)，可以接連出多少種不同的生成樹的問題了！

?

思路：Caylay定理，網(wǎng)上能找到證明，結(jié)果為n的n-2次方，然后利用快速二分冪求模去求解（以前總結(jié)的 快速二分冪求模）。

?code：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;
const int mod=2000000011;
int T,n;

ll big_mod(int a,int b) //關(guān)鍵實(shí)現(xiàn)函數(shù)，以前總結(jié)
{
ll ans=1,t=a;
while(b)
{
if(b&1) ans=ans*t%mod;
t=t*t%mod;
b>>=1;
}
return ans%mod;
}
int main()
{
scanf("%d",&T);
for (int ca=1;ca<=T;ca++)
{
int ans=1;
scanf("%d",&n);
if (n>1) ans=big_mod(n,n-2);
printf("Case #%d: %d\n",ca,ans);
}
}

總結(jié)

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