SMU Spring 2023 Trial Contest Round 9

A. Wrong Subtraction

 在k次操作里, 将n的最后一位数减1,如果是0就去掉,模拟一下就好了.

#include <bits/stdc++.h>
//#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 2e3 + 10, mod = 1e9 +7;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<vector<int>, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*
*/
vector<int> a, b;
void solve()
{
    string s;
    cin >> s >> m;
    while(m--){
        if(s.back() != '0'){
            s[s.size() - 1] --;
        }
        else
            s.erase(s.size() - 1, 1);
    }
    cout << s << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    //cin >>  Ke_scholar ;
    while(Ke_scholar--)
        solve();
    return 0;
}

B. Two-gram

暴力枚举,每次取两个字母再循环找一遍,输出最多次数的即可.

#include <bits/stdc++.h>
//#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 2e3 + 10, mod = 1e9 +7;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<vector<int>, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*
*/
vector<int> a, b;
void solve()
{
    cin >> n;
    string s;
    cin >> s;
    string ans ;
    int maxx = 0;
    for(int i = 0;i < s.size() - 1; i++){
        string s1 = s.substr(i,2);
        int x = 0;
        for(int j = 0; j <s.size() - 1; j ++){
            if(s.substr(j, 2) == s1)
                x ++;
        }
        if(x > maxx){
            maxx = x;
            ans = s1;
        }
    }
    cout << ans << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    //cin >>  Ke_scholar ;
    while(Ke_scholar--)
        solve();
    return 0;
}

C. Less or Equal

有个很坑的点,结束了看题解我才知道,k = 0 的时候分两种,如果最小的元素不是1的话应该输出1才对,害,我还以为0代表没有比较的,直接输出-1.

m == n 和 m == 0时特判一下,其余排个序再判断m前后是否相等的就行.

#include <bits/stdc++.h>
//#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 2e3 + 10, mod = 1e9 +7;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<vector<int>, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*
*/
vector<int> a, b;
void solve()
{
    cin >> n >> m;
    for(int i = 0;i < n; i++){
        int x;
        cin >> x;
        a.push_back(x);
    }
    sort(a.begin(), a.end());
    if(m == n){
        cout << a.back() << endl;
        return ;
    }
    if(m == 0){
        if(a.front() != 1)
            cout << 1 << endl;
        else
            cout << -1 << endl;
        return ;
    }
    if(a[m] == a[m - 1]){
        cout << -1 << endl;
    }
    else
        cout << a[m - 1] << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    //cin >>  Ke_scholar ;
    while(Ke_scholar--)
        solve();
    return 0;
}

D. Divide by three, multiply by two

将能被3整除的数或者是两倍关系的数的下标建成一条单向边,然后dfs去搜一条到达n的线路即可,至于为什么不用a[i]的值去建边,因为数据太大会寄.像这种.

input

2
1000000000000000000 3000000000000000000

output

3000000000000000000 1000000000000000000 

#include <bits/stdc++.h>
#define endl '\n'
#define int long long

using namespace std;
const int N = 110;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<vector<int>, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*
*/
bool vis[N];
int arr[N];
vector<int> v[N];
int a[N],ans[N];
void dfs(int val,int x){
    ans[x] = a[val];
    if(x >= n){
        for(int i = 1;i <= n; i++) {
            cout << ans[i] << ' ';
        }
        return ;
    }
    for(auto i : v[val])
        dfs(i, x + 1);
    return ;
}
void solve()
{
    cin >> n;
    for(int i = 0; i < n ;i++){
        cin >> a[i];
    }
    for(int i = 0; i < n; i++){
        for(int j = 0;j < n;j ++){
            if(i == j)
                continue;
            if(a[i] == a[j] * 3 || a[i] * 2 == a[j]){
                v[i].push_back(j);
            }
        }
    }
    for(int i = 0;i < n; i++){
        dfs(i, 1);
    }
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    //cin >> Ke_scholar;
    while(Ke_scholar--)
    {
        solve();
    }
    return 0;
}

E. Cyclic Components

将每条边的入度记录一下,两点的入度都是2的话就用并查集维护一下,判断是否成环.

#include <bits/stdc++.h>
#define endl '\n'
#define int long long

using namespace std;
const int N = 2e5 + 10;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<vector<int>, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*
*/
bool vis[N];
int arr[N];
int ans;
vector<PII> edge;
int sum[N];
int fa[N];
int find(int x){
    if(fa[x] == x)
        return x;
    return fa[x] = find(fa[x]);
}
void add(int x, int y){
    int dx = find(x);
    int dy = find(y);
    if(dx != dy)
        fa[dx] = dy;
    else
        ans ++;
}
void solve()
{
    cin >> n >> m;
    for(int i = 0; i < m; i++){
        int x, y;
        cin >> x >> y;
        edge.push_back({x, y});
        sum[x] ++;
        sum[y] ++;
    }
    for(int i = 1; i <= n; i++)
        fa[i] = i;
    for(int i = 0;i < m; i++){
        if(sum[edge[i].first] == 2 && sum[edge[i].second] == 2)
            add(edge[i].first, edge[i].second);
    }
    cout << ans << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    //cin >> Ke_scholar;
    while(Ke_scholar--)
    {
        solve();
    }
    return 0;
}

F. Consecutive Subsequence

用map做动态规划,记录最大的连续上升序列的结尾的值(记为maxid),然后从arr数组后面搜,如果等于maxid就放入ans里,maxid--,最后输出ans即可

#include <bits/stdc++.h>
#define endl '\n'
#define int long long

using namespace std;
const int N = 2e5 + 10;

//typedef long long ll;
typedef pair<int,int> PII;
//queue<PII> q1;
map<int, int > mp;
//priority_queue <int,vector<int>,greater<int> > q2;
int n,m,t,k;
/*
*/
bool vis[N];
int arr[N];
int ans;
int dp[N];
void solve()
{
    cin >> n;
    int maxx = 0, maxid;
    for(int i = 1; i <= n; i++){
        cin >> arr[i];
        mp[arr[i]] = max(mp[arr[i]], mp[arr[i] - 1] + 1);
        if(mp[arr[i]] > maxx) {
            maxx = mp[arr[i]];
            maxid = arr[i];
        }
    }
    cout << maxx << endl;
    vector<int> ams;
    for(int i = n; i > 0; i--){
        if(arr[i] == maxid){
            ams.push_back(i);
            maxid--;
        }
    }
    for(int i = ams.size() - 1; i >= 0; i--)
        cout << ams[i] << ' ';
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    int Ke_scholar = 1;
    //cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}