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。

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[2][250][250], ans;
int tem[250];
int n, k, tol;

string add (string a, string b) {
    string c;
    int s[1000] = {0};
    for (int i = 0; i < a.size(); i++) s[i] += a[i] - ‘0‘;
    for (int i = 0; i < b.size(); i++) s[i] += b[i] - ‘0‘;
    int len = max (a.size(), b.size() );
    for (int i = 0; i < len; ++i) {
        if (s[i] >= 10) {
            s[i + 1] += s[i] / 10, s[i] = s[i] % 10;
            if (i + 1 == len) len++;
        }
        c += ‘0‘ + s[i];
    }
    return c;
}
string operator * (string a, int k) {
    string c;
    int len = a.size(), x = 0;
    for (int i = 0, tem; i < len; ++i) {
        tem = (a[i] - ‘0‘) * k + x;
        c += ‘0‘ + tem % 10;
        x = tem / 10;
    }
    for (; x; x /= 10) c += ‘0‘ + x % 10;
    return c;
}
string operator * (string a, string b) {
    string c;
    int s[1000] = {0};
    for (int i = 0; i < a.size(); ++i)
        for (int j = 0; j < b.size(); ++j)
            s[i + j] += (a[i] - ‘0‘) * (b[j] - ‘0‘);
    int len = a.size() + b.size() - 1;
    for (int i = 0; i < len; ++i) {
        if (s[i] >= 10) {
            s[i + 1] += s[i] / 10, s[i] %= 10;
            if (i + 1 == len) len++;
        }
        c += ‘0‘ + s[i];
    }
    return c;
}
void make (int x) {
    tol = 0;
    for (int t = x; t <= n; t += 2) {
        tem[++tol] = t;
        if (t != n) tem[++tol] = t;
    }
    f[x - 1][0][0] = "1";
    string t;
    for (int i = 1; i <= tol; i++)
        for (int j = 0; j <= k; j++)
            if (tem[i] >= j) {
                if (j > 0) t = f[x - 1][i - 1][j - 1] * (tem[i] - j + 1);
                else       t = "0";
                f[x - 1][i][j] = add (f[x - 1][i - 1][j] , t );
            }
}
int main() {
    ios::sync_with_stdio (0);
    cin >> n >> k;
    make (1);
    make (2);
    ans = "0";
    for (int i = 0; i <= k; i++)
        ans = add (ans , f[1][tol][i] * f[0][2 * n - 1 - tol][k - i]);
    while (* (ans.end() - 1) == ‘0‘ && ans.size() > 1) ans.erase (ans.end() - 1);
    reverse (ans.begin(), ans.end() );
    cout << ans ;
}