SRM 585 DIV1 L2

记录dp(i, j)表示前i种卡片的排列，使得LISNumber为j的方法数。

#include <iostream>
#include <vector>
#include <string>
#include <string.h>
using namespace std;

typedef long long int64;

const int M = ;

int64 dpC[][];
int64 dpT[][];
int64 sum[];
int64 dp[][];

class LISNumber {
private:
    vector<int> num;
public:
    int64 f(int i, int j);
    int count(vector <int> cardsnum, int K);
};

int64 C(int m, int n) {
    if (m ==  || n ==  || m == n) {
        return ;
    }
    if (dpC[m][n] != -) {
        return dpC[m][n];
    }
    return dpC[m][n] = (C(m - , n) + C(m - , n - )) % M;
}

// m plate, n ball
int64 T(int m, int n) {
    // m >= 1
    if (m ==  || n == ) {
        return ;
    }
    if (dpT[m][n] != -) {
        return dpT[m][n];
    }
    return dpT[m][n] = (T(m - , n) + T(m, n - )) % M;
}

int64 LISNumber::f(int i , int j) {
    if (i == ) {
        if (j == num[]) {
            return dp[i][j] = ;
        } else {
            return dp[i][j] = ;
        }
    }
    if (j < num[i] || j > sum[i]) {
        return dp[i][j] = ;
    }
    if (dp[i][j] != -) {
        return dp[i][j];
    }

    // num[i] <= j <= sum[i]
    dp[i][j] = ;
    for (int k = ; k <= num[i]; k++) {
        dp[i][j] += (((C(j - k, num[i] - k) * T(sum[i] +  - j, k)) % M) * f(i - , j - k)) % M;
        dp[i][j] %= M;
    }
    return dp[i][j];
}

int LISNumber::count(vector <int> cardsnum, int K)
{
    num = cardsnum;
    memset(dp, -, sizeof(dp));
    memset(dpC, -, sizeof(dpC));
    memset(dpT, -, sizeof(dpT));
    memset(sum, , sizeof(sum));
    sum[] = cardsnum[];
    for (int i = ; i < cardsnum.size(); i++) {
        sum[i] = sum[i - ] + cardsnum[i];
    }
    return (int)f(cardsnum.size() - , K);
}