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)，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：
1 #pragma comment(linker, "/STACK:10240000,10240000")
2
3 #include <iostream>
4 #include <cstdio>
5 #include <algorithm>
6 #include <cstdlib>
7 #include <cstring>
8 #include <map>
9 #include <queue>
10 #include <deque>
11 #include <cmath>
12 #include <vector>
13 #include <ctime>
14 #include <cctype>
15 #include <set>
16 #include <bitset>
17 #include <functional>
18 #include <numeric>
19 #include <stdexcept>
20 #include <utility>
21
22 using namespace std;
23
24 #define mem0(a) memset(a, 0, sizeof(a))
25 #define mem_1(a) memset(a, -1, sizeof(a))
26 #define lson l, m, rt << 1
27 #define rson m + 1, r, rt << 1 | 1
28 #define define_m int m = (l + r) >> 1
29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
33 #define all(a) (a).begin(), (a).end()
34 #define lowbit(x) ((x) & (-(x)))
35 #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) {}
36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
38 #define pchr(a) putchar(a)
39 #define pstr(a) printf("%s", a)
40 #define sstr(a) scanf("%s", a)
41 #define sint(a) scanf("%d", &a)
42 #define sint2(a, b) scanf("%d%d", &a, &b)
43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
44 #define pint(a) printf("%d\n", a)
45 #define test_print1(a) cout << "var1 = " << a << endl
46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
48 #define mp(a, b) make_pair(a, b)
49 #define pb(a) push_back(a)
50
51 typedef unsigned int uint;
52 typedef long long LL;
53 typedef pair<int, int> pii;
54 typedef vector<int> vi;
55
56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
58 const int maxn = 1e8 + 17;
59 const int md = 1e9 + 7;
60 const int inf = 1e9 + 7;
61 const LL inf_L = 1e18 + 7;
62 const double pi = acos(-1.0);
63 const double eps = 1e-6;
64
65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
68 template<class T>T condition(bool f, T a, T b){return f?a:b;}
69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
70 int make_id(int x, int y, int n) { return x * n + y; }
71
72 struct ModInt {
73 static int MD;
74 int x;
75 ModInt(int xx = 0) { if (xx >= 0) x = xx % MD; else x = MD - (-xx) % MD; }
76 int get() { return x; }
77
78 ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
79 ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
80 ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
81 ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }
82
83 ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
84 ModInt operator -= (const ModInt &that) { x -= that.x; if (x < 0) x += MD; }
85 ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
86 ModInt operator /= (const ModInt &that) { *this = *this / that; }
87
88 ModInt inverse() const {
89 int a = x, b = MD, u = 1, v = 0;
90 while(b) {
91 int t = a / b;
92 a -= t * b; std::swap(a, b);
93 u -= t * v; std::swap(u, v);
94 }
95 if(u < 0) u += MD;
96 return u;
97 }
98
99 };
100 int ModInt::MD;
101 int p;
102 #define mint ModInt
103
104 mint C(mint fact[], mint fact_inv[], int n, int m) {
105 if (n < m) return 0;
106 if (n < p) return fact[n] * fact_inv[m] * fact_inv[n - m];
107 return C(fact, fact_inv, n / p, m / p) * C(fact, fact_inv, n % p, m % p);
108 }
109
110 mint get(mint a[], mint fact[], mint fact_inv[], int n, int m) {
111 if (n < 0 || m < 0) return 0;
112 if (n <= m) return a[n + 1] - 1;
113 mint ans = a[m + 1] - 1;
114 mint last = a[m];
115 rep_up1(i, n - m) {
116 int u = m + i;
117 last = last * 2 - C(fact, fact_inv, u - 1, m);
118 ans += last;
119 }
120 return ans;
121 }
122
123 int main() {
124 //freopen("in.txt", "r", stdin);
125 int a, b, c, d;
126 while (cin >> a >> b >> c >> d >> p) {
127 mint::MD = p;
128 mint x = 1;
129 mint mi2[100007], fact[100007], fact_inv[100007];
130 mi2[0] = fact[0] = fact_inv[0] = 1;
131 rep_up1(i, 100002) {
132 mi2[i] = mi2[i - 1] * 2;
133 fact[i] = fact[i - 1] * i;
134 fact_inv[i] = fact_inv[i - 1] / i;
135 }
136 cout << (get(mi2, fact, fact_inv, c, d) - get(mi2, fact, fact_inv, a - 1, d) -
137 get(mi2, fact, fact_inv, c, b - 1) + get(mi2, fact, fact_inv, a - 1, b - 1)).get() << endl;
138 }
139 return 0;
140 }
时间： 2024-10-12 20:42:27