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 Slines 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

---

Shahriar Manzoor

公式！別人的。   1 + (n, 2) + (n, 4)

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>

#define LEN 100

void Mult (int *a, int *b) {
    int c[LEN];
    memset (c, 0, sizeof (c));
    for (int i=0; i<LEN; ++i) {
        for (int k=0; i+k<LEN; ++k) {
            c[i + k] += a[i] * b[k];
        }
    }
    int car = 0;
    for (int i=0; i<LEN; ++i) {
        int t = car + c[i];
        c[i] = t % 10;
        car = t / 10;
    }
    assert (car == 0);
    memcpy (a, c, sizeof (c));
}

void Add (int *a, int *b) {
    int car = 0;
    for (int i=0; i<LEN; ++i) {
        int t = a[i] + b[i] + car;
        a[i] = t % 10;
        car = t / 10;
    }
    assert (car == 0);
}

void Put (int *a) {
    int i = LEN;
    while (--i > 0 && !a[i]) {}
    while (i >= 0) {
        printf ("%d", a[i--]);
    }
    printf ("\n");
}

void Dev (int *a, int d) {
    assert (0 < d && d < 10);
    int r = 0;
    for (int i=LEN-1; i>=0; --i) {
        int k = 10;
        while (--k > 0 && k * d > (r * 10 + a[i])) {}
        r = r * 10 + a[i] - k * d;
        assert (r < 10);
        a[i] = k;
    }
#ifdef _DEBUG
    if (r) {
        printf ("Warnning, not devisable!\n");
    }
#endif
}

void Sub (int *a, int *b) {
    int bor = 0; // borrow
    for (int i=0; i<LEN; ++i) {
        int t = a[i] - b[i] - bor;
        a[i] = (t + 10) % 10;
        bor = 1 - (t + 10) / 10;
    }
    assert (bor == 0);
}

void Get (int *a) {
    char s[100];
    if (scanf ("%s", s) == EOF) {
        exit (0);
    }
    memset (a, 0, sizeof (*a) * LEN);
    char *p = s + strlen (s);
    do {
        *a++ = *--p - '0';
    } while (p > s);
}

bool IsOne (int *a) {
    int one[LEN];
    memset (one, 0, sizeof (one));
    *one = 1;
    return memcmp (one, a, sizeof (one)) == 0;
}

bool IsTwo (int *a) {
    int two[LEN];
    memset (two, 0, sizeof (two));
    *two = 2;
    return memcmp (two, a, sizeof (two)) == 0;
}

bool IsZero (int *a) {
    int zero[LEN];
    memset (zero, 0, sizeof (zero));
    return memcmp (zero, a, sizeof (zero)) == 0;
}

int main () {
    int cs;
    scanf ("%d", &cs);
    while (cs--) {
        int n[LEN];
        Get (n);
        if (IsOne (n) || IsZero (n)) {
            printf ("1\n");
            continue;
        } else if (IsTwo (n)) {
            printf ("2\n");
            continue;
        }

        int a[LEN];
        memset (a, 0, sizeof (a));
        *a = 1;
        Mult (a, n);

        int b[LEN];
        memset (b, 0, sizeof (b));
        *b = 1;
        Mult (b, n);

        int c[LEN];
        memset (c, 0, sizeof (c));
        *c = 1;

        Sub (n, c);
        Mult (a, n);
        Mult (b, n);

        Dev (b, 2);

        Sub (n, c);
        Mult (a, n);

        Sub (n, c);
        Mult (a, n);

        Dev (a, 4);
        Dev (a, 6);

        Add (c, a);
        Add (c, b);

        Put (c);
    }
    return 0;
}