Acwing周赛102

倍增

这是一道简单数论题

using namespace std;
typedef long long LL;
const int N = 1e5 + 10;

int a[N], n;

int div(int x)
{
    if(x % 2 == 0)
        while(x % 2 == 0)    x /= 2;
    if(x % 3 == 0)
        while(x % 3 == 0)    x /= 3;
    return x;
}

int main()
{
    cin >> n;
    for(int i = 1; i <= n; ++ i)    scanf("%d", &a[i]);
    for(int i = 1; i <= n; ++ i)    a[i] = div(a[i]);
    for(int i = 1; i <= n; ++ i)
    {
        if(a[1] != a[i])
        {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
    return 0;
}

三元组

这道题目我们可以直接用stlmap套vector+stl的二分暴力来做也可以选择用前后缀分解，前后缀分解是我第一次见这种思想，会用一篇博客单独整理运用这种思想的题目

stl直接做

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
int a[N];
LL ans;
int main()
{
    int n, k;
    cin >> n >> k;

    map<LL, vector<int>>mp;
    for(int i = 1; i <= n; ++ i)
    {
        scanf("%d", &a[i]);
        mp[a[i]].push_back(i);
    }

    for(int i = 2; i < n; ++ i)
    {
        int ay = a[i];
        if(a[i] % k == 0)
        {
            LL ax = ay / k, az = 1LL * ay * k;
            if(mp.find(ax) != mp.end() && mp.find(az) != mp.end())
            {
                auto& v1 = mp[ax];
                auto& v2 = mp[az];
                int m = v2.size();
                int k1 = lower_bound(v1.begin(), v1.end(), i) - v1.begin();
                if(k1 >= 0)
                {
                    -- k1;
                    int k2 = upper_bound(v2.begin(), v2.end(), i) - v2.begin();
                    ans += 1LL * (k1 + 1) * (m - k2);
                }
            }
        }
    }
    cout << ans << endl;
    return 0;
}

前后缀分解去做

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
LL a[N];
map<LL, LL>l, r;
LL ans;
int main()
{
    LL n, k;
    scanf("%lld%lld", &n, &k);
    for(int i = 1; i <= n; ++ i)
    {
        scanf("%lld", &a[i]);
        r[a[i]] ++;
    }
    for(int i = 1; i <= n; ++ i)
    {
        r[a[i]] --;
        if(a[i] % k == 0)
        {
            if(l.find(a[i] / k) != l.end() && r.find(a[i] * k) != r.end())
                ans += 1LL * l[a[i] / k] * r[a[i] * k];
        }
        l[a[i]] ++;
    }
    cout << ans << endl;
}