codechef Count Relations(组合数 二项式定理)

题意

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}

Sol

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
*/

暴力

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;
}
/*
*/

原文地址：https://www.cnblogs.com/zwfymqz/p/9599352.html