[CodeForces 300D Painting Square]DP

http://codeforces.com/problemset/problem/300/D

题意：每一次操作可以选一个正方形，令边长为n，如果n为奇数那么可以从中间画一个十字，分成4个大小相等的边长为(n-1)/2的正方形。给一个正方形，求操作k次后能得到的不同图案的个数

思路：令f(s,k)表示边长为s的正方形操作k次后的答案总数，则f(s,k)=∑f(s/2,k₁)*f(s/2,k₂)*f(s/2,k₃)*f(s/2,k₄)，其中s为奇数，k₁+k₂+k₃+k₄=k-1，令g(s,k)=Σf(s,k₁)*f(s,k₂)，其中s为奇数，k1+k2=k。那么有f(s,k)=Σg(s/2,k')*g(s/2,k-1-k')，g(s,k)=Σf(s,k')*f(s,k-k')，其中s为奇数,k'取遍所有可能的值。由于s最多递归logs层，所以用一个不超过logs的数和k表示状态即可，然后递推计算，复杂度O(k²logs)。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + ;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

template<int mod>
struct ModInt {
    const static int MD = mod;
    int x;
    ModInt(ll x = ): x(x % MD) {}
    int get() { return x; }

    ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
    ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
    ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
    ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }

    ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
    ModInt operator -= (const ModInt &that) { x -= that.x; if (x < ) x += MD; }
    ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
    ModInt operator /= (const ModInt &that) { *this = *this / that; }

    ModInt inverse() const {
        int a = x, b = MD, u = , v = ;
        while(b) {
            int t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < ) u += MD;
        return u;
    }

};
typedef ModInt<> mint;

mint dp[][], f[][];

void pre_init() {
    for (int i = ; i <= ; i ++) dp[i][] = f[i][] = ;
    for (int i = ; i <= ; i ++) {
        dp[i][] = ;
        f[i][] = ;
        for (int j = ; j <= ; j ++) {
            for (int k = ; k <= j; k ++) {
                dp[i][j] += f[i - ][k] * f[i - ][j - k - ];
            }
            for (int k = ; k <= j; k ++) {
                f[i][j] += dp[i][k] * dp[i][j - k];
            }
        }
    }
}

int work(int n, int k) {
    if (k == ) return ;
    int p = ;
    while (n & ) {
        p ++;
        n = (n - ) / ;
    }
    if (n) p ++;
    return dp[p][k].get();
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int n, T, k;
    pre_init();
    while (cin >> T) {
        while (T --) {
            scanf("%d%d", &n, &k);
            printf("%d\n", work(n, k));
        }
    }
    return ;
}