1 介绍

本博客介绍容斥原理、概率和数学期望期望相关的题目。

2 训练

题目1：214Devu和鲜花

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 20, mod = 1e9 + 7;

LL A[N];
int down = 1;

int qmi(int a, int k, int p) {
    int res = 1;
    while (k) {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}

int C(LL a, LL b) {
    if (a < b) return 0;
    int up = 1;
    for (LL i = a; i > a - b; --i) up = i % mod * up % mod;
    
    return (LL)up * down % mod;
}

int main() {
    LL n, m;
    cin >> n >> m;
    for (int i = 0; i < n; ++i) cin >> A[i];
    
    for (int j = 1; j <= n - 1; j++) down = (LL)j * down % mod;
    down = qmi(down, mod - 2, mod);
    
    int res = 0;
    for (int i = 0; i < 1 << n; ++i) {
        LL a = m + n - 1, b = n - 1;
        int sign = 1;
        for (int j = 0; j < n; ++j) {
            if (i >> j & 1) {
                sign *= -1;
                a -= A[j] + 1;
            }
        }
        res = (res + C(a, b) * sign) % mod;
    }
    
    cout << (res + mod) % mod << endl;
    
    return 0;
}

题目2：215破译密码

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 50010;

int primes[N], cnt;
bool st[N];
int mobius[N], sum[N];

void init(int n) {
    mobius[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!st[i]) {
            primes[cnt++] = i;
            mobius[i] = -1;
        }
        for (int j = 0; primes[j] * i <= n; ++j) {
            int t = primes[j] * i;
            st[t] = true;
            if (i % primes[j] == 0) {
                mobius[t] = 0;
                break;
            }
            mobius[t] = mobius[i] * -1;
        }
    }
    
    for (int i = 1; i <= n; ++i) sum[i] = sum[i-1] + mobius[i];
}

int main() {
    init(N-1);
    
    int T;
    scanf("%d", &T);
    while (T--) {
        int a, b, d;
        scanf("%d%d%d", &a, &b, &d);
        a /= d, b /= d;
        int n = min(a, b);
        LL res = 0;
        for (int l = 1, r; l <= n; l = r + 1) {
            r = min(n, min(a / (a / l), b / (b / l)));
            res += (sum[r] - sum[l-1]) * (LL)(a / l) * (b / l);
        }
        
        printf("%lld\n", res);
    }
    
    return 0;
}

题目3：217绿豆蛙的归宿

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100010, M = 200010;

int n, m;
int h[N], e[M], w[M], ne[M], idx;
int dout[N];
double f[N];

void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

double dp(int u) {
    if (f[u] >= 0) return f[u];
    f[u] = 0;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        f[u] += (w[i] + dp(j)) / dout[u];
    }
    return f[u];
}

int main() {
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);
    
    for (int i = 0; i < m; ++i) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
        dout[a]++;
    }
    
    memset(f, -1, sizeof f);
    
    printf("%.2lf\n", dp(1));
    
    return 0;
}

题目4：218扑克牌

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 14;
const double INF = 1e20;

int A, B, C, D;
double f[N][N][N][N][5][5];

double dp(int a, int b, int c, int d, int x, int y) {
    double &v = f[a][b][c][d][x][y];
    if (v >= 0) return v;
    int as = a + (x == 0) + (y == 0);
    int bs = b + (x == 1) + (y == 1);
    int cs = c + (x == 2) + (y == 2);
    int ds = d + (x == 3) + (y == 3);
    if (as >= A && bs >= B && cs >= C && ds >= D) return v = 0;
    
    int sum = a + b + c + d + (x != 4) + (y != 4);
    sum = 54 - sum;
    if (sum <= 0) return v = INF;
    
    v = 1;
    if (a < 13) v += (13.0 - a) / sum * dp(a + 1, b, c, d, x, y);
    if (b < 13) v += (13.0 - b) / sum * dp(a, b + 1, c, d, x, y);
    if (c < 13) v += (13.0 - c) / sum * dp(a, b, c + 1, d, x, y);
    if (d < 13) v += (13.0 - d) / sum * dp(a, b, c, d + 1, x, y);
    if (x == 4) {
        double t = INF;
        for (int i = 0; i < 4; ++i) t = min(t, 1.0 / sum * dp(a, b, c, d, i, y));
        v += t;
    }
    if (y == 4) {
        double t = INF;
        for (int i = 0; i < 4; ++i) t = min(t, 1.0 / sum * dp(a, b, c, d, x, i));
        v += t;
    }
    
    return v;
}

int main() {
    cin >> A >> B >> C >> D;
    memset(f, -1, sizeof f);
    
    double t = dp(0, 0, 0, 0, 4, 4);
    if (t > INF / 2) t = -1;
    
    printf("%.3lf\n", t);
    
    return 0;
}