wannafly挑战赛29-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41650771/article/details/102711529

本文提供了两道算法竞赛题目的详细解答，包括数字标记验证和数论因子分解的优化方案，利用深度优先搜索和数论性质解决复杂问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A题目链接
直接求出所有可能出现的数字标记一下，然后看给出的数字是否全都出现过，然后在判断和是否是n
代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef __int128 LL;
map<ll , int> mp;
void dfs(ll x){
    if(x == 1){
        mp[x] = 1;
        return ;
    }
    if(!mp[x]){
        mp[x] = 1;
        dfs(x/2);
        dfs(x - x/2);
    }
}
int main()
{
    ll n,m;
    cin >> n >> m;
    LL sum = 0;
    dfs(n);
    int flag = 0;
    for(int i = 0; i < m ;i ++){
        ll x;
        cin >> x;
        if(!mp[x])flag = 1;
        sum += x;
    }
    if(!flag && sum == n) cout <<"misaka"<< endl;
    else cout <<"ham"<< endl;
}

B题目链接
在这里插入图片描述

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll f(ll x){
    ll sum = 1;
    while(x){
        sum = sum*(x%10);
        x /= 10;
    }
    return sum;
}
map<vector<ll> , ll> mp;
vector<ll> b(4),v(4);
int a[4] = {2,3,5,7};
int main()
{
    ll n,k;
    cin >> n >> k;
    if(k == 0){
        ll flag = 0;
        for(int i = 0; i < n; i ++){
            ll x;
            cin >> x;
            ll y = f(x);
            if(y == 1) flag ++;
        }
        
        cout << n*(n - 1)/2 - flag*(flag-1)/2 << endl;
    }
    else{
        ll flag = 0;
        ll cnt = 0;
        for(int i = 0; i < n; i ++){
            ll x;
            cin >> x;
            ll y = f(x);
            if(y){
                for(int j = 0; j < 4 ; j ++){
                    v[j] = 0;
                    while(y%a[j] == 0){
                        y /= a[j];
                        v[j] ++;
                    }
                    v[j] %= k;
                    b[j] = (k - v[j])%k;
                }
                cnt += mp[b];
                mp[v] ++;
            }
            else flag ++;
        }
        if(flag)
        cnt = flag*(flag - 1)/2 + cnt + flag*(n - flag);
        cnt = n*(n - 1)/2 - cnt;
        cout << cnt << endl;
    }
}

有一个加强版

#include <bits/stdc++.h>
using namespace std;
#define int long long 
const int  N = 2e5 + 4;
int a[200005];
map<std::vector<pair<int, int> > , int> mp;
signed  main()
{
	int n,k;
	cin >> n >> k;
	for(int i = 0; i < n; i ++){
		cin >> a[i];
	}
 
	int ans = 0;
	for(int i = 0 ; i < n ;i ++){
		std::vector<pair<int , int > > v,bu;
		int x = a[i];
		for(int j = 2; j*j <= x; j ++){
			int cnt = 0;
			if(x%j == 0){
				
				while(x%j == 0){
					x /= j;
					cnt ++;
				}
			}
			cnt %= k;
 
			if(cnt){
				v.push_back(make_pair(j,cnt));
				bu.push_back(make_pair(j,k - cnt));
			}
		}
		if(x > 1){
			v.push_back(make_pair(x, 1));
			bu.push_back(make_pair(x, k - 1));
		}
		//if(a[i] == 8) cout << v[0].first <<" " << v[0].second << endl;
		sort(v.begin(), v.end());
		sort(bu.begin(), bu.end());
		ans += mp[bu];
		mp[v] ++;
	}
	cout << ans << endl;
}