polya自主ac的第一道，討論方法：先把奇偶分開(1)順時針0度，90度，180度，270度 (2)鏡像豎線，水平線，兩條對角線。分別推出公式計算，實在推不出來，寫個模擬暴力找循環節，找規律，然后發現還要寫大數，就貼了個模板呢。。。后邊補了java版。ac代碼：

#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]; //可以控制大數的位數int len; //大數長度 public:BigNum(){ len = 1;memset(a,0,sizeof(a)); } //構造函數BigNum(const int); //將一個int類型的變量轉化為大數BigNum(const char*); //將一個字符串類型的變量轉化為大數BigNum(const BigNum &); //拷貝構造函數BigNum &operator=(const BigNum &); //重載賦值運算符，大數之間進行賦值運算friend istream& operator>>(istream&, BigNum&); //重載輸入運算符friend ostream& operator<<(ostream&, BigNum&); //重載輸出運算符BigNum operator+(const BigNum &) const; //重載加法運算符，兩個大數之間的相加運算BigNum operator-(const BigNum &) const; //重載減法運算符，兩個大數之間的相減運算BigNum operator*(const BigNum &) const; //重載乘法運算符，兩個大數之間的相乘運算BigNum operator/(const int &) const; //重載除法運算符，大數對一個整數進行相除運算BigNum operator^(const int &) const; //大數的n次方運算int operator%(const int &) const; //大數對一個int類型的變量進行取模運算bool operator>(const BigNum & T)const; //大數和另一個大數的大小比較bool operator>(const int & t)const; //大數和一個int類型的變量的大小比較void print(); //輸出大數 }; BigNum::BigNum(const int b) //將一個int類型的變量轉化為大數 {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) //將一個字符串類型的變量轉化為大數 {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) //拷貝構造函數 {int i;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n) //重載賦值運算符，大數之間進行賦值運算 {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) //重載輸入運算符 {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) //重載輸出運算符 {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 //兩個大數之間的相加運算 {BigNum t(*this);int i,big; //位數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 //兩個大數之間的相減運算 {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 //兩個大數之間的相乘運算 {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 //大數對一個整數進行相除運算 {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 //大數對一個int類型的變量進行取模運算 {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 //大數的n次方運算 {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 //大數和另一個大數的大小比較 {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 //大數和一個int類型的變量的大小比較 {BigNum b(t);return *this>b; }void BigNum::print() //輸出大數 {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);}} }

　　

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轉載于:https://www.cnblogs.com/RRRR-wys/p/8955669.html

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