Sgu 221.big Bishops (DP)

Question:

For an N * n (n <= 50) Board, place K Bishops (oblique walk) to find the total number of types that can be placed.

 

 

Solution:

The same as sgu 220, just add a high precision.

 

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 ;}

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Sgu 221.big Bishops (DP)