UVa 10213 How Many Pieces of Land?

Problem G
How Many Pieces of Land?
Input: Standard Input
Output: Standard Output
Time Limit: 3 seconds

 

You are given an elliptical shaped land and you are asked to choose n arbitrary points on its boundary. Then you connect all these points with one another with straight lines (that’s n*(n-1)/2 connections for n points). What is the maximum number of pieces of land you will get by choosing the points on the boundary carefully?

 

Fig: When the value of n is 6.

 

Input

The first line of the input file contains one integer S (0 < S < 3500), which indicates how many sets of input are there. The next S lines contain S sets of input. Each input contains one integer N (0<=N<2^31).

 

Output

For each set of input you should output in a single line the maximum number pieces of land possible to get for the value of N.

 

Sample Input:

4
1
2
3
4

 

Sample Output:

1
2
4
8

这个题在《入门经典》这本书里的做法是根据欧拉公式，V - E + F = 2。其中V是顶点数，包括所有顶点数和交点数，E是边数，包括椭圆弧和线段被交点分割成的段数，F是面数，包括题目所求和整个椭圆面这个大面。故题目所求 = E - V + 1。但是这样做一定会WA或TL，因为题目中对于输入的要求是3500个不大于2的31次方的数。因此这个题还有一个考点是高精度数的运算。同时网上有个本题的公式，C(n,4) + C(n,2) + 1。虽然无法推到和证明，但是还是可以理解这个公式的正确性的：

如果没有任何一个点，那么这个椭圆只有一个面——它本身。

每当有一条线（暂时不考虑相交），就会因为分割而多出一个面，n个点能组成C(n,2)条线，所以在原来的基础上加上这1 * C(n,2) = C(n,2)个面。

如果每条线都不与另外一条线相交，那么这样就是结果了，可惜的是，因为任意两点都要连线，故必定有相交，而每个相交又会导致多出一个面，而n个点有多少交线呢？n选2可以选出两个点构成一条线，n选4可以选择4个点，这四个点两两相连有且仅有一个交点（不考虑端点），因此n个点有1 * C(n,4) = C(n,4)个交点，导致多出C(n,4)个面。

因此最终的答案就是1 + C(n,2) + C(n,4)。

有了大数类，有了这个公式，这个题AC就不难了。如果像我一样结果看似“正确”但依然WA，就好好检查大数运算的类吧。提一下自己WA的原因：用Bign这个类的童鞋，这个公式是可以化简成(n * n - n * 5 + 18) * (n - 1) * n / 24 + 1的。但是我们写代码是不能这样写的，因为这个类无法减除负数来，所以当n在1-5之间的时候n * n - n * 5会计算出错误结果。改法就是写成n * n + 18 - n * 5……防止负数出现。

AC代码（0.132s）：

#include <iostream>#include <cstring>using namespace std;const int maxn = 200;struct Bign{    int s[maxn];    size_t len;        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] - '0';        return *this;    }        ///输出    string str() const    {        string res = "";        for (int i = 0; i < len; i++) {            res = (char)(s[i] + '0') + res;        }        if (res == "") {            res = "0";        }        return res;    }        ///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 (long 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 (long 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;    }        friend istream &operator>>(istream &is, Bign &num)    {        string s;        is >> s;        num = s.c_str();        return is;    }        friend ostream &operator<<(ostream &os, const Bign &num)    {        os << num.str();        return os;    }};int main(int argc, const char * argv[]) {    Bign n;    int M;    cin >> M;    while (M--) {        cin >> n;        cout << (n * n + 18 - n * 5) * (n - 1) * n / 24 + 1 << endl;    }        return 0;}