UVA 10519 !! Really Strange !!

大意：n个圆，两两相交于两点，把平面分成多少个区域。

思路：中学递推题，设已有n-1个圆，第n个圆与前n-1个圆形成2(n-1)交点，即多出2*(n-1)条弧，一条弧把平面分成2*(n-1)个平面。

所以：f(n) = f(n-1) + 2*(n-1)，然后根据f(2) - f(1) = 2; f(3) - f(2) = 4; f(n) - f(n-1) = 2*(n-1)可推出，f(n) = n*n - n + 2；

n=1时特判,WA3次。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAXN = 410;

struct bign
{
	int s[MAXN], len;
	bign () { memset(s, 0, sizeof(s)); len = 1;}
	bign (int num) { *this = num;}
	bign (const char *num) { *this = num;}
	bign operator = (const char *num)
	{
		len = strlen(num);
		for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
		return *this;
	}
	bign operator = (int num)
	{
		char s[MAXN];
		sprintf(s, "%d", num);
		*this = s;
		return *this;
	}
	bign operator + (const bign &b)
	{
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; g || i < max(len, b.len); i++)
		{
			int x = g;
			if(i < len) x += s[i];
			if(i < b.len) x += b.s[i];
			c.s[c.len++] = x % 10;
			g = x / 10;
		}
		return c;
	}
	void clean()
	{
		while(s[len-1] == 0 && len > 1) len--;
	}
	bign operator * (const bign &b)
	{
		bign c;
		c.len = len + b.len;
		for(int i = 0; i < len; i++)
		{
			for(int j = 0; j < b.len; j++)
			{
				c.s[i+j] += s[i] * b.s[j];
			}
		}
		for(int i = 0; i < c.len; i++)
		{
			c.s[i+1] += c.s[i] / 10;
			c.s[i] %= 10;
		}
		c.clean();
		return c;
	}
	bign operator - (const bign &b)
	{
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; i < len; i++)
		{
			int x = s[i]-g;
			if(i < b.len) x -= b.s[i];
			if(x >= 0) g = 0;
			else
			{
				g = 1;
				x += 10;
			}
			c.s[c.len++] = x;
		}
		c.clean();
		return c;
	}
	bool operator < (const bign &b)
	{
		if(len != b.len) return len < b.len;
		for(int i = len-1; i >= 0; i--)
		{
			if(s[i] != b.s[i]) return s[i] < b.s[i];
		}
		return 0;
	}
	bool operator > (const bign &b)
	{
		if(len != b.len) return len > b.len;
		for(int i = len-1; i >= 0; i--)
		{
			if(s[i] != b.s[i]) return s[i] > b.s[i];
		}
		return 0;
	}
	bool operator == (const bign &b)
	{
		return !(*this > b) && !(*this < b);
	}
	string str() const
	{
		string res = "";
		for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
		if(res == "") res = "0";
		return res;
	}
};

bign ans;

istream& operator >> (istream &in, bign &x)
{
	string s;
	in >> s;
	x = s.c_str();
	return in;
}

ostream& operator << (ostream &out, bign &x)
{
	out << x.str();
	return out;
}

int main()
{
	bign n;
	while(cin>>n)
	{
		if(n == 0) { printf("1\n"); continue;}
		ans = n*n - n + 2;
		cout<<ans<<endl;
	}
	return 0;
}