SGU 221.Big Bishops(DP)

题意：

给一个n*n(n<=50)的棋盘，放上k个主教（斜走），求能放置的种类总数。

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Solution :

同SGU 220，加个高精度就好了。

code

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

string  f[][][], ans;
int tem[];
int n, k, tol;

string add (string a, string b) {
    string c;
    int s[] = {};
    for (int i = ; i < a.size(); i++) s[i] += a[i] - '';
    for (int i = ; i < b.size(); i++) s[i] += b[i] - '';
    int len = max (a.size(), b.size() );
    for (int i = ; i < len; ++i) {
        if (s[i] >= ) {
            s[i + ] += s[i] / , s[i] = s[i] % ;
            if (i +  == len) len++;
        }
        c += '' + s[i];
    }
    return c;
}
string operator * (string a, int k) {
    string c;
    int len = a.size(), x = ;
    for (int i = , tem; i < len; ++i) {
        tem = (a[i] - '') * k + x;
        c += '' + tem % ;
        x = tem / ;
    }
    for (; x; x /= ) c += '' + x % ;
    return c;
}
string operator * (string a, string b) {
    string c;
    int s[] = {};
    for (int i = ; i < a.size(); ++i)
        for (int j = ; j < b.size(); ++j)
            s[i + j] += (a[i] - '') * (b[j] - '');
    int len = a.size() + b.size() - ;
    for (int i = ; i < len; ++i) {
        if (s[i] >= ) {
            s[i + ] += s[i] / , s[i] %= ;
            if (i +  == len) len++;
        }
        c += '' + s[i];
    }
    return c;
}
void make (int x) {
    tol = ;
    for (int t = x; t <= n; t += ) {
        tem[++tol] = t;
        if (t != n) tem[++tol] = t;
    }
    f[x - ][][] = "";
    string t;
    for (int i = ; i <= tol; i++)
        for (int j = ; j <= k; j++)
            if (tem[i] >= j) {
                if (j > ) t = f[x - ][i - ][j - ] * (tem[i] - j + );
                else       t = "";
                f[x - ][i][j] = add (f[x - ][i - ][j] , t );
            }
}
int main() {
    ios::sync_with_stdio ();
    cin >> n >> k;
    make ();
    make ();
    ans = "";
    for (int i = ; i <= k; i++)
        ans = add (ans , f[][tol][i] * f[][ * n -  - tol][k - i]);
    while (* (ans.end() - ) == '' && ans.size() > ) ans.erase (ans.end() - );
    reverse (ans.begin(), ans.end() );
    cout << ans ;
}