sgu292：Field for the Cemetery(结论+高精度)

题目大意：
$~~~~~~$一个$~q×c~$的矩阵，问最多能放入多少$~n×1~$的矩阵。

分析：
$~~~~~~$我们只考虑$~n\leq q,n\leq c~$的情况，其他的情况比较简单就不叙述了。
$~~~~~~$我们有最暴力的填法，使得最后空余的矩形为$~(q\mod n)×(c\mod n)~$；我们也可以先在矩形的周围围一圈，内部的矩形即为$~(n-2(q\mod n))×(n-2(c\mod n))~$，再暴力填，空余矩形即为$~(n-(q\mod n))×(n-(c\mod n))~$，两者中取答案最大值即可。
$~~~~~~$其实这道题是高精度的模板题。

AC code：

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <string>
#include <sstream>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#define pb push_back
#define mp make_pair
#define clr(a, b) memset(a, b, sizeof a)
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define per(i, a, b) for(int i = (a); i >= (b); --i)
typedef long long LL;
typedef double DB;
typedef long double LD;
using namespace std;

void open_init()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(0);
}

void close_file()
{
    #ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    #endif
}

const int MAX = 2009;
const int M = 10;

struct Bignum
{
    int a[MAX];
    Bignum() 
    {
        clr(a, 0);
        a[0] = 1;
    }
    Bignum(int k)
    {
        clr(a, 0);
        a[0] = 1;
        if(k)
        {
            a[0] = 0;
            while(k)
            {
                a[++a[0]] = k%M;
                k /= M;
            }
        }
    }
    Bignum& operator = (const Bignum &b)
    {
        clr(a, 0);
        memcpy(a, b.a, (b.a[0]+1)*sizeof(int));
        return *this;
    }
    inline void read()
    {
        char str[MAX] = "\0";
        scanf("%s", str);
        a[0] = strlen(str);
        per(i, a[0], 1)
            a[a[0]-i+1] = str[i-1]-‘0‘;
    }
    inline void adjust()
    {
        rep(i, 1, a[0])
        {
            while(a[i] < 0) a[i+1]--, a[i] += M;
            while(a[i] >= M) a[i+1]++, a[i] -= M; 
        }
        while(a[a[0]+1]) a[0]++;
        while(a[0] > 1 && !a[a[0]]) a[0]--;
    }
    inline void write()
    {
        per(i, a[0], 1)
            putchar(a[i]+‘0‘);
    }
}q, c, n, empty, ans;

inline bool operator < (const Bignum &a, const Bignum &b)
{
    if(a.a[0] != b.a[0]) return a.a[0] < b.a[0];
    per(i, a.a[0], 1)
        if(a.a[i] != b.a[i])
            return a.a[i] < b.a[i];
    return false;
}

inline bool operator == (const Bignum &a, const Bignum &b)
{
    if(a.a[0] != b.a[0]) return false;
    rep(i, 1, a.a[0])
        if(a.a[i] != b.a[i])
            return false;
    return true;
} 

inline bool operator > (const Bignum &a, const Bignum &b)
{
    if(a.a[0] != b.a[0]) return a.a[0] > b.a[0];
    per(i, a.a[0], 1)
        if(a.a[i] != b.a[i])
            return a.a[i] > b.a[i];
    return false;
}

inline bool operator <= (const Bignum &a, const Bignum &b)
{
    return !(a > b);
}

inline bool operator >= (const Bignum &a, const Bignum &b)
{
    return !(a < b);
}

inline Bignum operator + (const Bignum &a, const Bignum &b)
{
    Bignum ret;
    int len = max(a.a[0], b.a[0]);
    rep(i, 1, len)
        ret.a[i] = a.a[i]+b.a[i];
    ret.a[0] = len;
    ret.adjust();
    return ret;
}

inline void operator += (Bignum &a, const Bignum &b)
{
    a = a+b;
}

inline Bignum operator - (const Bignum &a, const Bignum &b)
{
    Bignum ret;
    int len = a.a[0];
    rep(i, 1, len)
        ret.a[i] = a.a[i]-b.a[i];
    ret.a[0] = len;
    ret.adjust();
    return ret;
}

inline void operator -= (Bignum &a, const Bignum &b)
{
    a = a-b;
}

inline Bignum operator * (const Bignum &a, int k)
{
    Bignum ret;
    ret.a[0] = a.a[0];
    rep(i, 1, a.a[0])
        ret.a[i] = a.a[i]*k;
    ret.adjust();
    return ret;
}

inline Bignum operator * (const Bignum &a, const Bignum &b)
{
    Bignum ret;
    per(i, a.a[0], 1)
        ret = ret*M+b*a.a[i];
    return ret;
}

template<class T>
inline Bignum operator *= (Bignum &a, const T &b)
{
    a = a*b;
}

inline Bignum operator / (const Bignum &a, const Bignum &b)
{
    Bignum ret, tmp;
    ret.a[0] = a.a[0];
    per(i, a.a[0], 1)
    {
        tmp = tmp*M+a.a[i];
        int l = 0, r = 9;
        while(l < r)
        {
            int mid = (l+r+1)>>1;
            if(b*mid > tmp) r = mid-1;
            else l = mid;
        }
        tmp -= b*l;
        ret.a[i] = l;
    }
    ret.adjust();
    return ret;
}

inline Bignum operator % (const Bignum &a, const Bignum &b)
{
    return a-a/b*b;
}

int main()
{
    open_init();

    q.read(), c.read(), n.read();
    if(q < n || c < n)
    {
        if(q < c) empty = c%n*q;
        else empty = q%n*c;
    }
    else
    {
        Bignum t1 = q%n, t2 = c%n;
        empty = min(t1*t2, (n-t1)*(n-t2));
    }
    ans = (q*c-empty)/n;
    ans.write();

    close_file();
    return 0;
}