[hdu5226]组合数求和取模(Lucas定理)

题意：给一个矩阵a，a[i][j] = C[i][j](i>=j) or 0(i < j)，求(x1,y1)，(x2,y2)这个子矩阵里面的所有数的和。

思路：首先问题可以转化为求(0,0)，(n,m)这个子矩阵的所有数之和。画个图容易得到一个做法，对于n<=m，答案就是2^0+2^1+...+2^m=2^(m+1)-1，对于n>m，答案由两部分构成，一部分是2^(m+1)-1，另一部分是sigma i:m+1->n f[i][m]，f[i][m]表示第i行前m列的数之和，f数组存在如下关系，f[i][m]=f[i-1][m]*2-C[i-1][m]，f[m][m]=2^m。还有另一种思路：第i列的所有数之和为C(i,i)+C(i+1,i)+...+C(n,i)=C(n+1,i+1)，于是答案就是sigma i:0->min(n,m) C(n+1,i+1)。

Lucas定理：由于题目给定的模是可变的质数，且质数可能很小，那么就不能直接用阶乘和阶乘的逆相乘了，需要用到Lucas定理，公式：C(n,m)%P=C(n/P,m/P)*C(n%P,m%P)%P，c(n,m)=0(n<m)。当然最终还是要预处理阶乘和阶乘的逆来得到答案。复杂度O(nlogP+nlogn)

下面是第一种思路的代码：

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef unsigned int uint;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e8 + 17;
const int md = 1e9 + 7;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

struct ModInt {
    static int MD;
    int x;
    ModInt(int xx = 0) { if (xx >= 0) x = xx % MD; else x = MD - (-xx) % 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 < 0) 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 = 1, v = 0;
        while(b) {
            int t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < 0) u += MD;
        return u;
    }

};
int ModInt::MD;
int p;
#define mint ModInt

mint C(mint fact[], mint fact_inv[], int n, int m) {
    if (n < m) return 0;
    if (n < p) return fact[n] * fact_inv[m] * fact_inv[n - m];
    return C(fact, fact_inv, n / p, m / p) * C(fact, fact_inv, n % p, m % p);
}

mint get(mint a[], mint fact[], mint fact_inv[], int n, int m) {
    if (n < 0 || m < 0) return 0;
    if (n <= m) return a[n + 1] - 1;
    mint ans = a[m + 1] - 1;
    mint last = a[m];
    rep_up1(i, n - m) {
        int u = m + i;
        last = last * 2 - C(fact, fact_inv, u - 1, m);
        ans += last;
    }
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int a, b, c, d;
    while (cin >> a >> b >> c >> d >> p) {
        mint::MD = p;
        mint x = 1;
        mint mi2[100007], fact[100007], fact_inv[100007];
        mi2[0] = fact[0] = fact_inv[0] = 1;
        rep_up1(i, 100002) {
            mi2[i] = mi2[i - 1] * 2;
            fact[i] = fact[i - 1] * i;
            fact_inv[i] = fact_inv[i - 1] / i;
        }
        cout << (get(mi2, fact, fact_inv, c, d) - get(mi2, fact, fact_inv, a - 1, d) -
                 get(mi2, fact, fact_inv, c, b - 1) + get(mi2, fact, fact_inv, a - 1, b - 1)).get() << endl;
    }
    return 0;
}