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;}