紫书 例题 10-23 UVa 10213（欧拉公式+高精度）

用欧拉公式V-E+F=2

V是顶点数，E是边数，F是面数

具体推导见https://blog.csdn.net/QWsin/article/details/53635397

要用高精度

#include<cstdio>

#include<algorithm>

#include<iostream>

#include<string>

#include<cstring>

#include<sstream>

#define REP(i, a, b) for(int i = (a); i < (b); i++)

using namespace std;

const int MAXN = 112;

struct bign

{

	int len, d[MAXN];

	void clean() 	{ while(len > 1 && !d[len-1]) len--; }   //去零

	bign() 			{ memset(d, 0, sizeof(d)); len = 1; }

	bign(int num) 	{ *this = num; }                         //bign a = 123（int）

	bign(char* num) { *this = num; }                         //bign a = "123"

	bign operator = (const char* num)                        //a = "123"

	{

		memset(d, 0, sizeof(d));

		len = strlen(num);

		REP(i, 0, len) d[i] = num[len-i-1] - '0';

		clean();

		return *this;

	}

	bign operator = (int num)                               //a = 123（int）

	{

		char s[20];

		sprintf(s, "%d", num);

		*this = s;

		return *this;

	}

	bign operator + (const bign& b)                        //加法

	{

		bign c = *this;	int i;

		for(i = 0; i < b.len; i++)

		{

			c.d[i] += b.d[i];

			if(c.d[i] > 9) c.d[i] %= 10, c.d[i+1]++;

		}

		while(c.d[i] > 9) c.d[i] %= 10, c.d[i+1]++;

		c.len = max(len, b.len) + 1;

		c.clean();

		return c;

	}

	bign operator - (const bign& b)                       //减法

	{

		bign c = *this; int i;

		for(i = 0; i < b.len; i++)

		{

			c.d[i] -= b.d[i];

			if(c.d[i] < 0) c.d[i] += 10, c.d[i+1]--;

		}

		while(c.d[i] < 0) c.d[i++] += 10, c.d[i]--;

		c.clean();

		return c;

	}

	bign operator * (const bign& b) const              //乘法，记得这里有const

	{

		bign c; c.len = len + b.len;

		REP(i, 0, len)

			REP(j, 0, b.len)

				c.d[i+j] += d[i] * b.d[j];

		REP(i, 0, c.len-1)

			c.d[i+1] += c.d[i] / 10, c.d[i] %= 10;

		c.clean();

		return c;

	}

	bign operator / (const bign& b)                    //除法

	{

		bign c = *this, a = 0;

		int i, j;

		for(i = len - 1; i >= 0; i--)

		{

			a = a * 10 + d[i];

			for(j = 0; j < 10; j++) if(a < b*(j+1)) break;

			c.d[i] = j;

			a = a - b * j;

		}

		c.clean();

		return c;

	}

	bign operator % (const bign& b)                    //模

	{

		bign a = 0;

		int i, j;

		for(i = len - 1; i >= 0; i--)

		{

			a = a * 10 + d[i];

			for(j = 0; j < 10; j++) if(a < b*(j+1)) break;

			a = a - b * j;

		}

		return a;

	}

	bign operator += (const bign& b)                 //+=

	{

		*this = *this + b;

		return *this;

	}

	bool operator <(const bign& b) const

	{

		if(len != b.len) return len < b.len;

		for(int i = len - 1; i >= 0; i--)

			if(d[i] != b.d[i])

				return d[i] < b.d[i];

		return false;

	}

	bool operator >(const bign& b)  const {return b < *this;}

	bool operator <=(const bign& b) const {return !(b < *this);}

	bool operator >=(const bign& b) const {return !(*this < b);}

	bool operator !=(const bign& b) const {b < *this || *this < b;}

	bool operator ==(const bign& b) const {return !(b < *this || *this < b);}

};

istream&  operator >> (istream& in, bign &x)

{

	int a;

    in>>a;

	x=a;

	return in;

}

ostream&  operator << (ostream& out,const bign &x)

{

	for(int i = x.len - 1; i >= 0; i--)

		printf("%d", x.d[i]);

    return out;

}

int main()

{

	int T;

	scanf("%d", &T);

	while(T--)

	{

		bign n;

		cin >> n;

		n = n * n * n * n + (bign)23 * n * n - bign(6) * n * n * n - (bign)18 * n;

		cout << n / 24 + (bign)1 << endl;

	}

	return 0;

}