[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e8 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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 = ) { if (xx >= ) 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 < ) 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;
     }

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

 mint C(mint fact[], mint fact_inv[], int n, int m) {
     if (n < m) return ;
     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 <  || m < ) return ;
     if (n <= m) return a[n + ] - ;
     mint ans = a[m + ] - ;
     mint last = a[m];
     rep_up1(i, n - m) {
         int u = m + i;
         last = last *  - C(fact, fact_inv, u - , 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 = ;
         mint mi2[], fact[], fact_inv[];
         mi2[] = fact[] = fact_inv[] = ;
         rep_up1(i, ) {
             mi2[i] = mi2[i - ] * ;
             fact[i] = fact[i - ] * i;
             fact_inv[i] = fact_inv[i - ] / i;
         }
         cout << (get(mi2, fact, fact_inv, c, d) - get(mi2, fact, fact_inv, a - , d) -
                  get(mi2, fact, fact_inv, c, b - ) + get(mi2, fact, fact_inv, a - , b - )).get() << endl;
     }
     return ;
 }