HackerRank

Good one.. marked as DP, so I was striking to work out a DP formula... well, not all DP have a formula - one type of DP is called memorized search: there's no formula, only previous calculated results can be reused..

I learnt from vivmbbs21's code:

#include <vector>
#include <iostream>
using namespace std;

#define MOD 1000000007

#define REP(i,s,e) for(int i = s; i < e; i ++)
vector<vector<int>> cr(302, vector<int>(302));

vector<vector<vector<int>>> memo(102, vector<vector<int>>(102, vector<int>(302)));
int solve(int d, int x, int m)
{
    if (x <= 0 || x > d) return 0;
    if (m == 0) return 1;
    if (memo[d][x][m] != -1) return memo[d][x][m];
    return memo[d][x][m] = (solve(d, x - 1, m - 1) + solve(d, x + 1, m - 1)) % MOD;
}

vector<vector<int>> memo2(10, vector<int>(302));
int solve2(int k, int n, int m, vector<int> &x, vector<int> &d)
{
    if (k == n) return m == 0 ? 1 : 0;
    if (memo2[k][m] != -1) return memo2[k][m];

    int ret = 0;
    REP(i, 0, m + 1) // at row k, for all possible steps to take
    {
        long long cret = 1LL * solve(d[k], x[k], i) * cr[m][i] % MOD;
        ret += cret * solve2(k + 1, n, m - i, x, d) % MOD;
        if (ret >= MOD) ret -= MOD;
    }
    return memo2[k][m] = ret;
}

int main()
{
    //    Pascal's Triangle
    REP(i, 0, 302)
    {
        cr[i][0] = cr[i][i] = 1;
        REP(j, 1, i) cr[i][j] = (cr[i - 1][j] + cr[i - 1][j - 1]) % MOD;
    }
    vector<vector<vector<int>>> init(102, vector<vector<int>>(102, vector<int>(302, -1)));
    memo = init;
    //
    int t; cin >> t;
    while (t--)
    {
        vector<vector<int>> init2(10, vector<int>(302, -1));
        memo2 = init2;

        int n, m; cin >> n >> m;
        vector<int> d(n), x(n);
        REP(i, 0, n) cin >> x[i];
        REP(i, 0, n) cin >> d[i];

        int ret = solve2(0, n, m, x, d);
        cout << ret << endl;
    }
}