UVA 10519 !! Really Strange !!

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=31329#problem/B

题意：

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=0时特判

大数模板。

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 210;

char numstr[maxn];

struct bign
{
int len, s[maxn];

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

bign(int num)
{
*this = num;
}

bign(const char* num)
{
*this = num;
}

bign operator = (const int num)
{
char s[maxn];
sprintf(s, "%d", num);
*this = s;
return *this;
}

bign operator = (const char* num)
{
len = strlen(num);
for (int i = 0; i < len; i++) s[i] = num[len - i - 1] & 15;
return *this;
}

///输出
const char* str() const
{
if (len)
{
for (int i = 0; i < len; i++)
numstr[i] = '0' + s[len - i - 1];
numstr[len] = '\0';
}
else strcpy(numstr, "0");
return numstr;
}

///去前导零
void clean()
{
while (len > 1 && !s[len - 1]) len--;
}

///加
bign operator + (const bign& b) const
{
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;
}

///减
bign operator - (const bign& b) const
{
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;
}

///乘
bign operator * (const bign& b) const
{
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 - 1; i++)
{
c.s[i + 1] += c.s[i] / 10;
c.s[i] %= 10;
}
c.clean();
return c;
}

///除
bign operator / (const bign &b) const
{
bign ret, cur = 0;
ret.len = len;
for (int i = len - 1; i >= 0; i--)
{
cur = cur * 10;
cur.s[0] = s[i];
while (cur >= b)
{
cur -= b;
ret.s[i]++;
}
}
ret.clean();
return ret;
}

///模、余
bign operator % (const bign &b) const
{
bign c = *this / b;
return *this - c * b;
}

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

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

bign operator += (const bign &a)
{
*this = *this + a;
return *this;
}

bign operator -= (const bign &a)
{
*this = *this - a;
return *this;
}

bign operator *= (const bign &a)
{
*this = *this * a;
return *this;
}

bign operator /= (const bign &a)
{
*this = *this / a;
return *this;
}

bign operator %= (const bign &a)
{
*this = *this % a;
return *this;
}
};

int main()
{
bign n;
while (gets(numstr))
{
n = numstr;
puts(n == 0 ? "1" : (n * n - n + 2).str());//在计算的过程中如果出现负数，此模板不可处理
}
return 0;
}

Java

import java.math.BigInteger;
import java.util.Scanner;

public class Main{
public static void main(String[] args )
{
Scanner cin = new Scanner(System.in);

while( cin.hasNext())
{
BigInteger n = cin.nextBigInteger();
if( n.equals(BigInteger.valueOf(0)))
{
System.out.println("1");
}
else
{
n = n.multiply(n).add(n.negate()).add(BigInteger.valueOf(2));
System.out.println(n);
}
}
}
}