HDU 4365Palindrome graph(找规律 快速幂取模)

Palindrome graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1306    Accepted Submission(s): 398

Problem 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<n<=10000,0<=m<=2000,0<k<=1000000), indicate n*n grid and k kinds of pigments.
Then follow m lines,for each line,there are 2 integer i,j.indicated that grid(i,j) (0<=i,j<n) has been filled with color.
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
   
   
   
   

 

Author

FZU

 

Source

2012 Multi-University Training Contest 7

 

题目大意：flip是翻转的意思，rotate是旋转的意思。就是说给你一个n*n的格子，然后每个格子图上任意颜色，但是通过翻转旋转必须一样。高度轴对称和中心对称。

 解题思路：经过思考可以发现，我们将正方形先等分成四份，如果一份确定，其它三个必然确定。而一个小正方形中是否存在依赖关系呢，答案是肯定的。所以再将正方形细分为两个三角形即可，一个三角形确定，小正方形确定，大正方形也确定了。这是总数sum。而有一些是涂色了的，个数便会减少，但是可以经过变换等价的点只能计算一次，详见代码。

题目地址：Palindrome graph

一直以为是mod=1e9+7，这个题目竟然是1e8+7。 题目得看清！！

AC代码：

#include<iostream>
#include<cstdio>
#include<map>
using namespace std;
int n,len;
const int mod=1e8+7;  //一直写的是1e9+7
map<int,int> mp;

void conver(int &x,int &y)
{
    while(1)
    {
        if(x<=len&&y<=len) break;
        int tmp=x;
        x=y;
        y=n+1-tmp;
    }
    if(x<y) swap(x,y);
}

__int64 fastmi(__int64 base,__int64 p)
{
    __int64 ans=1;
    while(p)
    {
        if(p&1)
            ans=(ans*base)%mod;
        base=(base*base)%mod;
        p>>=1;
    }
    return ans;
}

int main()
{
    int m,k,i;
    int x,y;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        mp.clear();
        len=(n+1)>>1;
        int sum;
        if(n&1) sum=(n/2+1)*(n/2+2)/2;
        else sum=(n/2)*(n/2+1)/2;

        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            x++,y++;
            conver(x,y);
            x=10000*x+y;
            if(!mp[x])
            {
                mp[x]=1;
                sum--;
            }
        }

        printf("%I64d\n",fastmi(k,sum));
    }
    return 0;
}