题面

洛谷

题解

\(f[i][j]\)表示有i个人参与游戏，从庄家（即1）数j个人获胜的概率是多少

\(f[1][1] = 1\)

这样就可以不用讨论淘汰了哪些人和顺序

枚举选庄家选那张牌, 枚举下一次的庄家

可以得到这次的庄家

然后转移即可

Code

#include<bits/stdc++.h>

#define LL long long

#define RG register

using namespace std;

template<class T> inline void read(T &x) {

    x = 0; RG char c = getchar(); bool f = 0;

    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;

    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();

    x = f ? -x : x;

    return ;

}

template<class T> inline void write(T x) {

    if (!x) {putchar(48);return ;}

    if (x < 0) x = -x, putchar('-');

    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;

    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;

}

const int N = 55;

int a[N];

double f[N][N];

int main() {

    //freopen(".in", "r", stdin);

    //freopen(".out", "w", stdout);

    int n, m;

    read(n); read(m);

    for (int i = 1; i <= m; i++) read(a[i]);

    f[1][1] = 100;

    for (int i = 2; i <= n; i++)

        for (int j = 1; j <= m; j++) {

            int p = a[j]%i == 0 ? i : a[j]%i;

            for (int k = 1; k < i; k++) {

                if (++p > i) p = 1;

                f[i][p] += f[i-1][k]/m;

            }

        }

    for (int i = 1; i <= n; i++)

        printf("%.2lf%% ", f[n][i]);

    return 0;

}