Blocks
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 720 |
|
Accepted: 201 |
Description
Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.
Input
The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.
Output
For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.
Sample Input
2
1
2
Sample Output
2
6
Source
給定一塊有n個點(diǎn)的木塊，用四種顏色涂色，其中兩種顏色只能用偶數(shù)次，求有多少種涂色方法。
一看就知是生成函數(shù)，可惜從沒用過。小試身手，沒想到竟然弄出來了。結(jié)果應(yīng)該是對的，就是不知過程是不是可以這樣寫。
設(shè)四種顏色分別為w，x，y，z，其中y，z只能用偶數(shù)次，我的推導(dǎo)過程如下：
最后得到的公式是(2^( n - 1 ))(2^(n-1)+1)
注意到10007是素?cái)?shù),由費(fèi)爾馬定理,可以先把n-1mod（10007-1）,減小計(jì)算量,剩下的就是快速取冪了.
#include <iostream>
using namespace std;
const int mod=10007;
int pow(int n)
{
if(n==0)
return 1;
if(n&1)
{
return (pow(n-1)<<1)%mod;
}
else
{
int temp=pow(n>>1);
return (temp*temp)%mod;
}
}
int main(int argc, char *argv[])
{
int t,n,temp;
cin>>t;
while(t--)
{
cin>>n;
temp=pow((n-1)%(mod-1));
cout<<(temp*(temp+1))%mod<<endl;
}
return 0;
}
//由于近日POJ登不上，上面的代碼未曾提交過