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ACM快速幂模板HDU4365

作者：qm38dal | 来源：互联网 | 2023-10-10 08:37
ACM快速幂模板HDU4365快速幂模板:lintquick(lintbase,lintpow){lintret1;while(pow){if(pow&1)ret(ret*ba
ACM 快速幂模板 

HDU4365

//快速幂模板:
lint quick(lint base, lint pow)
{lint ret&＃61;1;while(pow){if (pow&1) ret&＃61;(ret*base)%mod;base&＃61;(base*base)%mod;pow>>&＃61;1;}return ret%mod;//记得最后mod&＃xff0c;比如mod&＃61;&＃61;1的情况
}

原题&＃xff1a;

题解&＃xff1a;http://www.cnblogs.com/bo-tao/archive/2012/08/15/2640380.html

/** Author: NICK WONG* Created Time: 2015/8/5 21:27:00* File Name: e.cpp*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define out(x) cout<<#x<<": "<const double eps(1e-8);
const int maxn&＃61;10100;
const long long maxint&＃61;-1u>>1;
const long long maxlong&＃61;maxint*maxint;
typedef long long lint;
set > s;
int n,m;
lint k;
const long long mod&＃61;100000007;void init()
{
}void change(int& x, int& y)
{if (x>n/2) x&＃61;n&＃43;1-x;if (y>n/2) y&＃61;n&＃43;1-y;if (x>y) swap(x,y);
}lint quick(lint base, lint pow)
{lint ret&＃61;1;while(pow){if (pow&1) ret&＃61;(ret*base)%mod;base&＃61;(base*base)%mod;pow>>&＃61;1;}return ret;
}void work()
{lint ans&＃61;0;s.clear();for (int i&＃61;1; i<&＃61;(n&＃43;1)/2; i&＃43;&＃43;)ans&＃43;&＃61;i;for (int i&＃61;1; i<&＃61;m; i&＃43;&＃43;){int x,y;cin>>x>>y;x&＃43;&＃43;; y&＃43;&＃43;;change(x,y);s.insert(make_pair(x,y));}ans-&＃61;s.size();ans&＃61;quick(k,ans);cout<}int main()
{while(cin>>n>>m>>k){init();work();}return 0;
}



E - Palindrome graph
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4365


Description

In addition fond of programing, Jack also loves painting. He likes to draw many interesting graphics on the paper. 
One day,Jack found a new interesting graph called Palindrome graph. No matter how many times to flip or rotate 90 degrees, the palindrome graph are always unchanged. 
Jack took a paper with n*n grid and K kinds of pigments.Some of the grid has been filled with color and can not be modified.Jack want to know:how many ways can he paint a palindrome graph?

Input

There are several test cases. 
For each test case,there are three integer n m k(0 Then follow m lines,for each line,there are 2 integer i,j.indicated that grid(i,j) (0<&＃61;i,j You can suppose that jack have at least one way to paint a palindrome graph.

Output

For each case,print a integer in a line,indicate the number of ways jack can paint. The result can be very large, so print the result modulo 100 000 007.

Sample Input

3 0 2
4 2 3
1 1
3 1

Sample Output

8
3






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