HDU 4365Palindrome graph(找规律 快速幂取模)
Palindrome graphTime 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 易做图.
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 易做图 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; }
补充:软件开发 , C++ ,