HDU1812 - Count the Tetris

polya自主ac的第一道，讨论方法：先把奇偶分开(1)顺时针0度，90度，180度，270度 (2)镜像竖线，水平线，两条对角线。分别推出公式计算，实在推不出来，写个模拟暴力找循环节，找规律，然后发现还要写大数，就贴了个模板呢。。。后边补了java版。ac代码：

#include<iostream>
#include<string>
#include <cstdio>
#include <cstring>
#include<iomanip>
#include<algorithm>
using namespace std;#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4class BigNum
{
private:int a[500];    //可以控制大数的位数int len;       //大数长度
public:BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数BigNum(const int);       //将一个int类型的变量转化为大数BigNum(const char*);     //将一个字符串类型的变量转化为大数BigNum(const BigNum &);  //拷贝构造函数BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算BigNum operator^(const int  &) const;    //大数的n次方运算int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较void print();       //输出大数
};
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{int c,d = b;len = 0;memset(a,0,sizeof(a));while(d > MAXN){c = d - (d / (MAXN + 1)) * (MAXN + 1);d = d / (MAXN + 1);a[len++] = c;}a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{int t,k,index,l,i;memset(a,0,sizeof(a));l=strlen(s);len=l/DLEN;if(l%DLEN)len++;index=0;for(i=l-1;i>=0;i-=DLEN){t=0;k=i-DLEN+1;if(k<0)k=0;for(int j=k;j<=i;j++)t=t*10+s[j]-'0';a[index++]=t;}
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{int i;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{int i;len = n.len;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = n.a[i];return *this;
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{char ch[MAXSIZE*4];int i = -1;in>>ch;int l=strlen(ch);int count=0,sum=0;for(i=l-1;i>=0;){sum = 0;int t=1;for(int j=0;j<4&&i>=0;j++,i--,t*=10){sum+=(ch[i]-'0')*t;}b.a[count]=sum;count++;}b.len =count++;return in;}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{int i;cout << b.a[b.len - 1];for(i = b.len - 2 ; i >= 0 ; i--){cout.width(DLEN);cout.fill('0');cout << b.a[i];}return out;
}BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{BigNum t(*this);int i,big;      //位数big = T.len > len ? T.len : len;for(i = 0 ; i < big ; i++){t.a[i] +=T.a[i];if(t.a[i] > MAXN){t.a[i + 1]++;t.a[i] -=MAXN+1;}}if(t.a[big] != 0)t.len = big + 1;elset.len = big;return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算
{int i,j,big;bool flag;BigNum t1,t2;if(*this>T){t1=*this;t2=T;flag=0;}else{t1=T;t2=*this;flag=1;}big=t1.len;for(i = 0 ; i < big ; i++){if(t1.a[i] < t2.a[i]){j = i + 1;while(t1.a[j] == 0)j++;t1.a[j--]--;while(j > i)t1.a[j--] += MAXN;t1.a[i] += MAXN + 1 - t2.a[i];}elset1.a[i] -= t2.a[i];}t1.len = big;while(t1.a[len - 1] == 0 && t1.len > 1){t1.len--;big--;}if(flag)t1.a[big-1]=0-t1.a[big-1];return t1;
}BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{BigNum ret;int i,j,up;int temp,temp1;for(i = 0 ; i < len ; i++){up = 0;for(j = 0 ; j < T.len ; j++){temp = a[i] * T.a[j] + ret.a[i + j] + up;if(temp > MAXN){temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);up = temp / (MAXN + 1);ret.a[i + j] = temp1;}else{up = 0;ret.a[i + j] = temp;}}if(up != 0)ret.a[i + j] = up;}ret.len = i + j;while(ret.a[ret.len - 1] == 0 && ret.len > 1)ret.len--;return ret;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{BigNum ret;int i,down = 0;for(i = len - 1 ; i >= 0 ; i--){ret.a[i] = (a[i] + down * (MAXN + 1)) / b;down = a[i] + down * (MAXN + 1) - ret.a[i] * b;}ret.len = len;while(ret.a[ret.len - 1] == 0 && ret.len > 1)ret.len--;return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算
{int i,d=0;for (i = len-1; i>=0; i--){d = ((d * (MAXN+1))% b + a[i])% b;}return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{BigNum t,ret(1);int i;if(n<0)exit(-1);if(n==0)return 1;if(n==1)return *this;int m=n;while(m>1){t=*this;for( i=1;i<<1<=m;i<<=1){t=t*t;}m-=i;ret=ret*t;if(m==1)ret=ret*(*this);}return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{int ln;if(len > T.len)return true;else if(len == T.len){ln = len - 1;while(a[ln] == T.a[ln] && ln >= 0)ln--;if(ln >= 0 && a[ln] > T.a[ln])return true;elsereturn false;}elsereturn false;
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{BigNum b(t);return *this>b;
}void BigNum::print()    //输出大数
{int i;cout << a[len - 1];for(i = len - 2 ; i >= 0 ; i--){cout.width(DLEN);cout.fill('0');cout << a[i];}cout << endl;
}BigNum q_pow(BigNum a,int b) {BigNum ans = BigNum(1);while(b) {if(b&1)ans = (ans*a);a = (a*a);b>>=1;}return ans;
}
int M, n;
int main()
{while(~scanf("%d%d",&n,&M)) {int cc=0;BigNum m = BigNum(M),ans = BigNum(0);int t;if(n%2==0){t = n*n;ans = ans + q_pow(m,t);t = n*n/4;ans = ans + q_pow(m,t);t = n*n/2;ans = ans + q_pow(m,t);t = n*n/4;ans = ans + q_pow(m,t);t = n*n/2;ans = ans + BigNum(2)*q_pow(m,t);t = (n*n-n)/2+n;ans = ans + BigNum(2)*q_pow(m,t);}else{t = n*n;ans = ans + q_pow(m,t);t = (n+1)*(n-1)/4 + 1;ans = ans + q_pow(m,t);t = (n+1)*(n-1)/2 + 1;ans = ans + q_pow(m,t);t = (n+1)*(n-1)/4 + 1;ans = ans + q_pow(m,t);t = (n*n+n)/2;ans = ans + BigNum(4)*q_pow(m,t);}ans = ans / 8;ans.print();}return 0;
}

Java版

import java.math.*;
import java.util.*;
import java.io.*;
public class Main
{public static void main(String[] args){Scanner cin = new Scanner(System.in);while (cin.hasNext()){int n = cin.nextInt(),t;BigInteger ans=BigInteger.valueOf(0),m = cin.nextBigInteger();if(n%2==0) {ans = ans.add(m.pow(n*n));ans = ans.add(m.pow(n*n/4));ans = ans.add(m.pow(n*n/4));ans = ans.add(m.pow(n*n/2));ans = ans.add(BigInteger.valueOf(2).multiply(m.pow((n*n)/2)));ans = ans.add(BigInteger.valueOf(2).multiply(m.pow((n*n+n)/2)));}else {ans = ans.add(m.pow(n*n));ans = ans.add(m.pow((n+1)*(n-1)/4+1));ans = ans.add(m.pow((n+1)*(n-1)/2+1));ans = ans.add(m.pow((n+1)*(n-1)/4+1));ans = ans.add(BigInteger.valueOf(4).multiply(m.pow((n*n+n)/2)));}ans = ans.divide(BigInteger.valueOf(8));System.out.println(ans);}}
}