标签：等于 turn eve play mod 暴力 include 组合数 src
求有多少元素属于$1 \sim N$的集合满足
R1 = {（x，y）：x和y属于B，x不是y的子集，y不是x的子集，x和y的交集等于空集}
R2 = {（x，y）：x和y属于B，x不是y的子集，y不是x的子集，x和y的交集不等于空集}
神仙题啊Orz
我整整推了两个小时才推出来
首先写暴力
/* */ #include<iostream> #include<cstdio> #include<cstring> //#define int long long #define LL int const int MAXN = 1001; using namespace std; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } int N; int C[MAXN][MAXN], Po2[MAXN]; main() { cin >> N; Po2[0] = 1; for(int i = 1; i <= 1000; i++) Po2[i] = 2 * Po2[i - 1]; C[0][0] = 1; for(int i = 1; i <= 1000; i++) { C[i][0] = 1; for(int j = 1; j <= i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } int ans = 0; for(int i = 1; i <= N - 1; i++) { int now = C[N][i], sum = 0; for(int j = 1; j <= N - i; j++) sum += C[N - i][j]; //ans += now * sum; ans += now * (Po2[N - i] - 1); } cout << ans / 2 << " "; ans = 0; for(int i = 2; i <= N - 1; i++) { int res1 = C[N][i], sum1 = 0; for(int j = 1; j <= i - 1; j++) { int res2 = C[i][j], sum2 = 0; for(int k = 1; k <= N - i; k++) { sum2 += C[N - i][k]; } sum1 += res2 * sum2; } ans += res1 * sum1; } cout << ans / 2; return 0; } /* 100 50 10000006 */
然后一层一层展开即可
最后的答案为
第一问：
$$\frac{3^n + 1}{2} - 2^n$$‘
第二问：
$$\frac{-3 * 3^n + 4^n - 1}{2} + 3 * 2^{n - 1}$$
/* */ #include<iostream> #include<cstdio> #include<cstring> #define int long long #define LL int const int MAXN = 1001, mod = 100000007, inv = 50000004; using namespace std; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } int N; int C[MAXN][MAXN], Po2[MAXN], Po3[MAXN], Po4[MAXN]; int f(int a, int p) { int base = 1; while(p) { if(p & 1) base = (base * a) % mod; a = (a * a) % mod; p >>= 1; } return base % mod; } main() { int T; cin >> T; while(T--) { cin >> N; /*Po2[0] = Po3[0] = Po4[0] = 1; for(int i = 1; i <= 1000; i++) Po2[i] = 2 * Po2[i - 1], Po3[i] = 3 * Po3[i - 1], Po4[i] = 4 * Po4[i - 1]; int ans = (Po3[N] + 1) / 2 - Po2[N]; cout << ans << " "; ans = 0; ans = (-3 * Po3[N] + Po4[N] - 1) / 2; ans += Po2[N - 1] * 3; cout << ans; */ //cout << f(2, mod - 2) % mod; int ans = ((f(3, N) + 1) * inv - f(2, N) + mod) % mod; cout << ans << " "; ans = ((-3 * f(3, N) + mod) % mod + f(4, N) - 1 + mod) * inv + f(2, N - 1) * 3 % mod; ans %= mod; cout << ans << endl; } return 0; } /* */
codechef Count Relations(组合数 二项式定理)
标签：等于 turn eve play mod 暴力 include 组合数 src
原文地址：https://www.cnblogs.com/zwfymqz/p/9599352.html
