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;
}