题目链接：
HDU 1812 Count the Tetris
题意：
有一个n*n的网格，用m种颜色来涂，求本质不同的方案数？
分析；
Polya原理+高精度。
从原题中描述可以看出要考虑旋转，但是翻转的情况好像没有说清吧。。。
还有需要记录已经算出来的答案，当再次询问时，直接输出，否则会TLE，这也太恶心了吧。。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cassert>
#include <ctime>
using namespace std;
typedef long long ll;const int BASE = 1000;
const int SIZE = 510;int vis[35][35];struct BigInteger{int length, digit[SIZE];BigInteger(int number = 0) :length(number) {assert(0 <= number && number <= BASE);memset(digit, 0, sizeof(digit));digit[0] = number;}BigInteger normalize() {while(length && digit[length - 1] == 0) length--;return *this;}int operator [](int index) const {return digit[index];}int& operator [](int index) {return digit[index];}void output() {printf("%d", digit[length - 1]);for(int i = length - 2; i >= 0; i--) {printf("%03d", digit[i]);}printf("\n");}
}p[35][35];inline bool operator < (const BigInteger& a, const BigInteger& b)
{if(a.length != b.length) return a.length < b.length;for(int i = 0; i < a.length; ++i) {if(a[i] != b[i]) return a[i] < b[i];}return false;
}inline BigInteger operator + (const BigInteger& a, const BigInteger& b)
{BigInteger c;c.length = max(a.length, b.length) + 1;for(int i = 0, carry = 0; i < c.length; ++i) {carry += a[i] + b[i];c[i] = carry % BASE;carry /= BASE;}return c.normalize();
}inline BigInteger operator - (const BigInteger& a, int b)
{assert(0 <= b && b < BASE);BigInteger c;c.length = a.length;for(int i = 0, carry = -b; i < a.length; ++i) {carry += a[i];c[i] = carry;if(c[i] < 0) {c[i] += BASE;carry = -1;}else carry = 0;}return c.normalize();
}inline BigInteger operator * (const BigInteger& a, const BigInteger& b)
{BigInteger c;c.length = a.length + b.length;for(int i = 0; i < a.length; ++i) {for(int j = 0, carry = 0; j <= b.length; ++j) {carry += a[i] * b[j] + c[i + j];c[i + j] = carry % BASE;carry /= BASE;}}return c.normalize();
}inline BigInteger operator / (const BigInteger& a, int b)
{assert(0 <= b && b < BASE);BigInteger c;c.length = a.length;for(int i = c.length - 1, carry = 0; i >= 0; i--) {carry = carry * BASE + a[i];c[i] = carry / b;carry %= b;}return c.normalize();
}inline BigInteger operator ^ (const BigInteger& a, int b)
{BigInteger res = 1, tmp = a;while(b) {if(b & 1) res = res * tmp;tmp = tmp * tmp;b >>= 1;}return res;
}inline void solve(int n, int m)
{if(vis[n][m]) {p[n][m].output();return ;}vis[n][m] = 1;BigInteger ans = 0, tmp = m;ans.normalize();tmp.normalize();ans = ans + (tmp ^ (n * n)) + (tmp ^ ((n * n + 3) / 4)) * 2 + (tmp ^ ((n * n + 1) / 2));//考虑翻转和旋转if(n & 1) {ans = ans + (tmp ^ ((n * n - n) / 2 + n)) * 2; //沿中间线翻转ans = ans + (tmp ^ ((n * n - n) / 2 + n)) * 2; //沿对角线翻转}else {ans = ans + (tmp ^ (n * n / 2)) * 2; //沿中间线翻转ans = ans + (tmp ^ ((n * n - n) / 2 + n)) * 2; //沿对角线翻转}ans = ans / 8;ans.output();p[n][m] = ans;
}int main()
{int n, m;memset(vis, 0, sizeof(vis));while(~scanf("%d%d", &n, &m)) {solve(n, m);}return 0;
}