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

題意：

大意：n個圓，兩兩相交于兩點(diǎn)，把平面分成多少個區(qū)域。

思路：中學(xué)遞推題，設(shè)已有n-1個圓，第n個圓與前n-1個圓形成2(n-1)交點(diǎn)，即多出2*(n-1)條弧，一條弧把平面分成2*(n-1)個平面。

所以：f(n) = f(n-1) + 2*(n-1)，然后根據(jù)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時特判

大數(shù)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;}///輸出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;}///去前導(dǎo)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;}///減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());//在計算的過程中如果出現(xiàn)負(fù)數(shù)，此模板不可處理}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);}}} }

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