2024.3.04~2024.3.10 by manjuan

給你一個陣列 a1,a2…an 。請計算有多少個圖元 (i,j,k,l)符合以下條件：
· \(1\) \(\le\) \(i\)<\(j\)<\(k\)\(<\)\(l\)\(\le\)n
· a\(i\)\(=\) a\(k\) 和 a\(j\) \(=\) a\(l\)

\(Input\)
The first line contains a single integer \(t\) (\(1\)≤\(t\)≤\(100\)) — the number of test cases.

The first line of each test case contains a single integer \(n\) (\(4\)≤\(n\)≤\(3000\)) — the size of the array a.

The second line of each test case contains n integers a\(1\),a\(2\),…,a\(n\)(\(1\)≤a\(i\)≤\(n\)) — the array a.

It's guaranteed that the sum of n in one test doesn't exceed 3000.

\(Output\)
For each test case, print the number of described tuples.

題目連結:https://codeforces.com/contest/1400/problem/D

雜湊一下其實可以更快,但是資料量不大,n方解決即可;

AC程式碼:

#include"bits/stdc++.h"
using namespace std;
using ll=long long;
#define int long long
const int Mod=998244353;


void save_the_people(){
    int n,ans{0};
    cin>>n;
    vector<int> a(n);
    for(int i{0};i<n;i++){
        cin>>a[i];
        a[i]--;
    }
    vector<int> c1(n),c2(n);
    for(int i{0};i<n;i++){
        fill(c1.begin(), c1.end(),0);
        fill(c2.begin(), c2.end(),0);
        for(int l{i+1};l<n;l++){
            c2[a[l]]++;
        }
        int res{0};
        for(int k{i+1};k<n;k++){
            res-=c1[a[k]];
            c2[a[k]]--;
            if(a[i]==a[k])
                ans+=res;
            res+=c2[a[k]];
            c1[a[k]]++;
        }
    }
    cout<<ans<<"\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t{1};
    cin>>t;
    while(t--) {
        save_the_people();
    }

}

題目連結:https://codeforces.com/contest/1934/problem/D1

一道位運算的題,只要找出初始數n轉為二進位制後第一二個1的位置和目標數第一個1的位置的關係即可解決;

/***
 *    author:  touirst
 *    created: now
***/
#include <bits/stdc++.h>

using namespace std;

using ll=long long;
const int N=1e9+5;

void save_the_newnew() {
    ll n,m;
    cin>>n>>m;
    //cout<<n<<"\n";
    //cout<<(n^m)<<"\n";
    if((n^m)<n){cout<<1<<"\n"<<n<<" "<<m<<"\n";return;}
    ll lim{0},bn{63},bm{63};
    vector<int> a(65),b(65);
    //cout<<n<<"\n";
    while(!((n>>bn)&1)){
        bn--;
    }
    //cout<<bn<<"\n";
    bn--;
    //cout<<bn<<"\n";
    while(!((n>>bn)&1)){
        bn--;
    }
    //cout<<bn<<"\n";
    lim=(1ll<<(bn+1))-1;
    //cout<<lim<<"\n";
    while(!((m>>bm)&1)){
        bm--;
    }
    if(lim==m){
        cout<<1<<"\n"<<n<<" "<<m<<"\n";return;
    }else if(lim<m||lim>=n){
        cout<<"-1\n";return;
    }else{
        cout<<2<<"\n"<<n<<" "<<lim<<" "<<m<<"\n";
    }
    //cout<<bm<<"\n";
    //cout<<((7>>0)&1)<<" "<<((7>>1)&1)<<" "<<((7>>2)&1);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int ttt{1};
    cin>>ttt;
    while(ttt--){
        save_the_newnew();
    }
}