THE SAMS' CONTEST

Problem 0

BACKGROUND

Raju has recently passed BSc. Engineering in Computer Science & Engineering from BUET ( Bangladesh University of Extraordinary Talents), the best university of Bangladesh. After passing, he has been appointed manager of BCS ( Bangladesh Calculation
Society ). His first project is to visit a district and then to report the total number of distinct regions there. But, going there, he is astonished to see a very strange phenomena of the local people. He discovered that, the people make their area circular.
Every two area have exactly two points to intersect and no three circles intersect in a common point. There are so many people so that it's very hard to calculate total number of distinct regions for Raju. So, only in this case he seeks for your help.

Input

The problem contains multiple test cases. You are given total number of circular area n in separate lines (n<=10¹⁰⁰).
And you know the number houses in the world can't be fraction or negative. Input is terminated by end of file.

Output

For each line of the input, your correct program should output the value of the total number of regions in separate lines for each value of n.

Sample Input

3
4

Sample Output

8
14

题意：给定n，输出n个圆最多能把一平面分割成几块

思路：规律题，注意特判0的情况

代码：

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
const int N = 2222;
const int MAXBIGN = 305;


struct bign {
    int s[MAXBIGN];
    int len;
    bign() {
		len = 1;
		memset(s, 0, sizeof(s));
    }
	
    bign operator = (const char *number) {
		len = strlen(number);
		for (int i = 0; i < len; i++)
			s[len - i - 1] = number[i] - '0';
		return *this;
    }
    bign operator = (const int num) {
		char number[N];
		sprintf(number, "%d", num);
		*this = number;
		return *this;
    }
	
    bign (int number) {*this = number;}
    bign (const char* number) {*this = number;}
	
    bign operator + (const bign &c){  
		bign sum;
		int t = 0;
		sum.len = max(this->len, c.len);
		for (int i = 0; i < sum.len; i++) {
			if (i < this->len) t += this->s[i];
			if (i < c.len) t += c.s[i];
			sum.s[i] = t % 10;
			t /= 10;
		}
		
		while (t) {
			sum.s[sum.len++] = t % 10;
			t /= 10;
		}
		
		return sum;  
    }
	
	bign operator * (const bign &c){  
		bign sum; bign zero;
		if (*this == zero || c == zero)
			return zero;
		int i, j;
		sum.len = this->len + c.len;
		for (i = 0; i < this->len; i++) {
			for (j = 0; j < c.len; j ++) {
				sum.s[i + j] += this->s[i] * c.s[j];
			}
		}
		for (i = 0; i < sum.len; i ++) {
			sum.s[i + 1] += sum.s[i] / 10;
			sum.s[i] %= 10;
		}
		sum.len ++;
		while (!sum.s[sum.len - 1]) {
			sum.len --;
		}
		return sum;  
    }	
    bign operator - (const bign &c) {
		bign ans;
		ans.len = max(this->len, c.len);
		int i;
		
		for (i = 0; i < c.len; i++) {
			if (this->s[i] < c.s[i]) {
				this->s[i] += 10;
				this->s[i + 1]--;
			}
			ans.s[i] = this->s[i] - c.s[i];
		}
		
		for (; i < this->len; i++) {
			if (this->s[i] < 0) {
				this->s[i] += 10;
				this->s[i + 1]--;
			}
			ans.s[i] = this->s[i];
		}
		while (ans.s[ans.len - 1] == 0) {
			ans.len--;
		}
		if (ans.len == 0) ans.len = 1;
		return ans;
    }
	
    void put() {
		if (len == 1 && s[0] == 0) {
			printf("0");
		} else {
			for (int i = len - 1; i >= 0; i--)
				printf("%d", s[i]);
		}
    }
	
    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& b) const { return !(b != *this); }
};

char a[105];

int main() {
	while (~scanf("%s", a)) {
		bign n = a;
		bign zero = 0;
		bign one = 1;
		bign ans = 2;
		if (n == zero)
			printf("1");
		else {
			ans = n * (n - one) + ans;
			ans.put();
		}
		printf("\n");
	}
	return 0;
}