Codeforces 300D Painting Square dp

Painting Square

转换一下变成 a 层的完全四叉树， 占领 k 个点有多少种方案， 点能被占当且仅当它的父亲被占。

a <= 30， 所以我们把每层都dp出来， dp[ i ][ j ] 表示 i 层完全四叉树占领 k 个点的方案数。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 7340033;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

int dp[31][1007];
int q, n, k;

int main() {
    dp[0][0] = 1;
    for(int o = 1; o < 31; o++) {
        dp[o][0] = 1;
        dp[o][1] = 1;
        for(int c = 0; c < 4; c++) {
            for(int i = 1000; i >= 1; i--) {
                for(int j = 1; i + j <= 1000; j++) {
                    add(dp[o][i + j], 1LL * dp[o][i] * dp[o - 1][j] % mod);
                }
            }
        }
    }
    scanf("%d", &q);
    while(q--) {
        scanf("%d%d", &n, &k);
        int c = 0;
        while((n & 1) && n != 1) {
            c++;
            n--;
            n >>= 1;
        }
        printf("%d\n", dp[c][k]);
    }
    return 0;
}

/*
*/