AC了两道题，但是排名靠后了好多，大家都说是手速场qwq，可惜我不会dfs，第三题就放了，不然估计能混到250左右？Maybe~

1.AcWing 4209. 三元组

原代码：

#include<bits/stdc++.h>
using namespace std;
int a[110],b[110],c[110];
int x, y, z;
int main()
{
    int n;
    cin >> n;
    for (int i = 0;i < n;i ++)
    {
        cin >> a[i] >> b[i] >> c[i];
    }
    for (int i = 0;i < n;i ++)
    {
        x += a[i];
        y += b[i];
        z += c[i];
    }
    if (x == 0 && y == 0 & z == 0) cout << "YES";
    else cout << "NO";
    return 0;
}

写的略傻，不必开三个数组，完全可以这么写：PS：开始用大模板写题了算了，比赛再用吧，补题不用

#include <bits/stdc++.h>
using namespace std;

int x, y, z;

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

    int n;
    cin >> n;

    while (n --)
    {
        int a,b,c;
        cin >> a >> b >> c;
        x += a;
        y += b;
        z += c;
    }

    if (!x && !y&& !z) cout << "YES";
    else cout << "NO";

    return 0;
}

2.AcWing 4210. 数字

原代码：

#include<bits/stdc++.h>
using namespace std;
int sum;

int gcd(int sum, int t)
{
    if (t != 0) return gcd(t, sum % t);
    return sum;
}

int main()
{
    int n;
    cin >> n;
    for (int i = 2;i <= n-1;i ++)
    {
        int s = n;
        while (s > 0)
        {
            int mod = s % i;
            s /= i; 
            sum += mod;
        }
    }

    int t = n-2;

    int tem = gcd(sum, t);
    sum /= tem;
    t /= tem;

    cout << sum << '/' << t;

    return 0;
}

补题：

#include<bits/stdc++.h>
using namespace std;
int sum;

int gcd(int a, int b)
{
    if (a % b == 0) return b;
    return gcd(b, a % b);
}

int main()
{
    int n;
    cin >> n;

    for (int i = 2;i < n;i ++)
    {
        int s = n;
        while (s > 0)
        {
            sum += s % i;
            s /= i; 
        }
    }

    int t = n - 2;
    int tem = gcd(sum, t);
    sum /= tem;
    t /= tem;

    cout << sum << '/' << t;

    return 0;
}

或者直接调用库函数的最大公约数 __gcd()：

#include<bits/stdc++.h>
using namespace std;
int sum;

int main()
{
    int n;
    cin >> n;

    for (int i = 2;i < n;i ++)
    {
        int s = n;
        while (s > 0)
        {
            sum += s % i;
            s /= i; 
        }
    }

    int t = n - 2;
    int tem = __gcd(sum, t);
    sum /= tem;
    t /= tem;

    cout << sum << '/' << t;

    return 0;
}

3.AcWing 4211. 序列重排

可证明合法序列中任一元素的2因子和3因子之和唯一，故可只双关键字排序即可。

补题代码：

#include<bits/stdc++.h>
using namespace std;

const long long N = 105;

pair<int, long long> f[N];

int main()
{
    int n;
    cin >> n;
    for (int i = 0;i < n;i ++) cin >> f[i].second;

    for (int i = 0;i < n;i ++)
    {
        long long x = f[i].second;
        while (x % 2 == 0) {f[i].first ++; x /= 2;}
        while (x % 3 == 0) {f[i].first --; x /= 3;}
    }
    sort(f, f+n);

    for (int i = 0;i < n;i ++) cout << f[i].second << ' ';

    return 0;
}

y总的三关键字vector排序：

#include<bits/stdc++.h>
using namespace std;

const int N = 105;

vector<long long> f[N];

int get(long long a, int b)
{
    int sum = 0;
    while (a % b == 0) {sum ++; a /= b;}
    return sum;
}

int main()
{
    int n;
    cin >> n;
    for (int i = 0;i < n;i ++)
    {
        long long x;
        cin >> x;
        f[i] = {get(x,2), ~get(x,3), x};
    }

    sort(f, f+n);

    for (int i = 0;i < n;i ++) cout << f[i][2] << ' ';

    return 0;
}

struct结构体写法，多关键字的结构体排序需要手写cmp：

#include<bits/stdc++.h>
using namespace std;

const int N = 105;

struct code
{
    int x;
    int y;
    long long z;
}f[N];

int get(long long a, int b)
{
    int sum = 0;
    while (a % b == 0) {sum ++; a /= b;}
    return sum;
}

bool cmp(code m, code n)
{
    if (m.x == n.x) return m.y > n.y;
    return m.x < n.x;
}

int main()
{
    int n;
    cin >> n;
    for (int i = 0;i < n;i ++)
    {
        long long x;
        cin >> x;
        f[i].x = get(x,2);
        f[i].y = get(x,3);
        f[i].z = x;
        // 或者 f[i] = {get(x,2), get(x,3), x};
    }

    sort(f, f+n, cmp);

    for (int i = 0;i < n;i ++) cout << f[i].z << ' ';

    return 0;
}

借鉴dalao的dfs做法

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 110;

int n;

LL a[N], b[N];
bool f[N];

void dfs(int u)
{
    if (u == n + 1)
    {
        for (int i = 1;i <= n;i ++)
        {
            cout << b[i] << ' ';
        }
        return;
    }
    for (int i = 0;i < n;i ++)
    {
        if (u == 1)
        {
            b[u] = a[i];
            f[i] = true;
            dfs (u + 1);
            f[i] = false; // 状态回溯
            b[u] = 0;
        }
        else if ((a[i] == b[u - 1] * 2 ||a[i] * 3 == b[u - 1]) && !f[i])
        {
            b[u] = a[i];
            f[i] = true;
            dfs (u + 1);
            f[i] = false;
            b[u] = 0;
        }
    }
}

int main()
{
    cin >> n;
    for (int i = 0;i < n;i ++) cin >> a[i];

    dfs(1);

    return 0;
}