Simulate the hundred prisoner problem

A very famous “ Counter intuitive ” Intelligence questions _ Bili, Bili _bilibili A very famous “ Counter intuitive ” Intelligence questions ：—— Yes 100 The names and numbers are 1,2,3......100 Our prisoners .—— The number of each prisoner is written on a piece of paper . this 100 A note , Randomly placed in a room 100 In a box with the same appearance . Put a piece of paper in each box .—— Each prisoner can enter the room alone once , View at will 50 Boxes , Cover the box after reading , No trace left . If he can just see “ Own number ”, Deemed to have completed the task , Need to leave the room immediately .—— Prisoners can discuss strategies before starting , Video playback volume 608827、 The amount of bullet screen 964、 Number of likes 18316、 The number of coins dropped 1722、 Number of collectors 12315、 Forwarding number 5070, Video author lovelyff998, Author's brief introduction , Related video ： Mathematics has a fatal flaw , Where should you stand , To live to the end ？, Infinite chocolate paradox , American competition questions , The foreigner hasn't been right for a long time , Chinese students can do it in one move , Ten thousand words ： How far has today's game technology developed ？, How to explain the grandfather paradox ？, people class low quality The amount double translate Jian Appreciation （14）, Why successful people never write to-do lists （To-Do List） ？► The method she mentioned is more effective - Millions of popular online entrepreneurs Rachel Hollis Rachel ‧ Holly （ Chinese and English subtitles ）, Can feel “ Energy ” Mind and body games , The biggest illusion in the world https://www.bilibili.com/video/BV1kt4y1t75F?p=1&share_medium=android&share_plat=android&share_session_id=55f1ff0e-fda4-43be-8c5a-112768033498&share_source=QQ&share_tag=s_i×tamp=1656936860&unique_k=Ru3JMdE&vd_source=78c5a56b2b804449168dbad7346576d9

#include <iostream>
#include <time.h>
using namespace std;
const int N = 100;
int dfs(int a[], int x, int vis[]) {
    if (vis[a[x]]) {
        return 0;
    }
    vis[a[x]] = 1;
    return dfs(a, a[a[x]], vis) + 1;
}
int main() {
    srand(time(0));
    int cnt = 0, T = 100000;
    for (int round = 0; round < T; round++) {
        int a[128] = {0};
        for (int i = 1; i <= N; i++) {
            a[i] = i;
        }
        for (int i = N; i > 1; i--) {
            int r = 1 + rand() % i;
            int t = a[i];
            a[i] = a[r];
            a[r] = t;
        }
        int vis[128] = {0};
        for (int i = 1; i <= N; i++) {
            if (dfs(a, i, vis) > (N >> 1)) {
                cnt++;
                break;
            }
        }
    }
    cout << 1.0 * cnt / T;
    return 0;
}