SMU Summer 2023 Contest Round 5

SMU Summer 2023 Contest Round 5

A. Points in Segments

\(\mathcal{O}(n \times m)\) 做法数据范围小,直接把每次的\(l - r\)跑一遍标记一下,最后跑一遍循环统计哪些没有被标记的并且输出就好了

#include <bits/stdc++.h>
#define int long long

using namespace std;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n,m;
    cin >> n >> m;
    vector<int> a(m + 1);
    int ans = 0;
    for(int i = 0;i < n;i ++){
        int x,y;
        cin >> x >> y;
        for(int j = x;j <= y;j ++){
            a[j] = 1;
        }
    }

    for(int i = 1;i <=m;i ++ ){
        ans += (a[i] == 0);
    }

    if(!ans){
        cout << 0 << endl;
    }else{
        cout << ans << endl;
        for(int i = 1;i <= m;i ++)
            if(!a[i])
                cout << i << ' ';
    }

    return 0;
}

\(\mathcal{O}(n + m)\) 做法数据再大点就可以开个前缀和数组,让每次输入的\(l\)所在的值为1,\(r\)的值为-1,然后跑一遍前缀和,统计为0的个数就行

#include <bits/stdc++.h>

using namespace std;

int main() {

	int n, m;
	cin >> n >> m;
	vector<int> cnt(m + 2);
	for (int i = 0; i < n; ++i) {
		int l, r;
		cin >> l >> r;
		++cnt[l];
		--cnt[r + 1];
	}
	for (int i = 1; i <= m; ++i)
		cnt[i] += cnt[i - 1];

	vector<int> ans;
	for (int i = 1; i <= m; ++i) {
		if (cnt[i] == 0)
			ans.push_back(i);
	}

	cout << ans.size() << endl;
	for (auto it : ans) cout << it << " ";
	cout << endl;

	return 0;
}

B. Obtaining the String

\(\mathcal{O}(n^3)\) 长度只有50,直接跑暴力了,每次对比s和t是否相同,不同就去s后面找到当前与t相同的,然后从后一个一个交换过来,判断s能不能变成b就看s里每个字符数量和b是不是一样

#include <bits/stdc++.h>
#define int long long

using namespace std;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n;
    cin >> n;
    string s,t;
    cin >> s >> t;

    if(s == t){
        cout << 0 << endl;
        return 0;
    }

    vector<int> numa(100),numb(100);
    for(int i = 0;i < s.size();i ++){
        numa[s[i] - 'a']++;
        numb[t[i] - 'a']++;
    }

    for(int i = 0;i <= 100;i ++){
        if(numa[i] != numb[i]){
            cout << -1 << endl;
            return 0;
        }
    }

    int ans = 0;
    vector<int> res;
    for(int i = 0;i < n - 1;i ++){
        if(s[i] == t[i]) continue;
        for(int j = i + 1;j < n;j ++){
            if(s[j] == t[i]){
                for(int k = j;k > i;k--){
                    swap(s[k],s[k - 1]);
                    ans++;
                    res.push_back(k);
                }
                break;
            }
        }
    }

    cout << ans << endl;
    for(auto i : res)
        cout << i << ' ';

    return 0;
}

\(\mathcal{O}(n^2)\) 做法就是在后面找到了之后单独拿出来从后跑一遍

#include <bits/stdc++.h>

using namespace std;

int main() {

	int n;
	string s, t;
	cin >> n >> s >> t;

	vector<int> ans;
	for (int i = 0; i < n; ++i) {
		if (s[i] == t[i]) continue;
		int pos = -1;
		for (int j = i + 1; j < n; ++j) {
			if (s[j] == t[i]) {
			        pos = j;
			        break;
			}
		}
		if (pos == -1) {
			cout << -1 << endl;
			return 0;
		}
		for (int j = pos - 1; j >= i; --j) {
			swap(s[j], s[j + 1]);
			ans.push_back(j);
		}
	}

	cout << ans.size() << endl;
	for (auto it : ans) cout << it + 1 << " ";
	cout << endl;

	return 0;
}

C. Songs Compression

\(\mathcal{O}(nlog_2n)\) 先把所有歌都压缩,如果还是小于\(m\)的话直接输出-1,否则就将内存的差值从小到大排序,每次从最小的差值开始加,看看能还原多少首歌,最后用总数去减掉还原的就是压缩的个数了

#include <bits/stdc++.h>
#define int long long

using namespace std;

typedef pair<int,int> PII;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n,m;
    cin >> n >> m;

    vector<PII> a(n);
    for(auto& [x,y] : a)
        cin >> x >> y;

    sort(a.begin(),a.end(),[](PII x,PII y){
        return x.first - x.second < y.first - y.second;
    });

    int ans = 0, now = 0;
    for(auto i : a){
        now += i.second;
    }
    if(now > m){
        cout << -1 << endl;
        return 0;
    }

    for(auto i : a){
        if(i.first - i.second + now <= m){
            now += i.first - i.second;
            ans ++;
        }else
            break;
    }
    cout << n - ans << endl;

    return 0;
}

D. Walking Between Houses

\(\mathcal{O}(k)\) 首先\(s < k\)和\(s < (n - 1) \times k\)这两种情况肯定是不行的,至于为什么,可以自己推导一下,然后就是一个贪心的思想,每次走当前能走的最远距离,即每次能走的距离为\(min(n-1,s-(k-1))\),然后s也要跟着更新

#include <bits/stdc++.h>
#define int long long

using namespace std;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n,k,s;
    cin >> n >> k >> s;

    if(s > (n - 1) * k || s < k){
        cout << "NO" << endl;
        return 0;
    }

    cout << "YES" << endl;
    int sum = 0, now = 1;
    for(; k;k-- ){
        int i = min(s - k + 1, n - 1);
        s -= i;
        if( i + now > n){
            cout << now - i << ' ';
            now -= i;
        }else {
            cout << now + i << ' ';
            now += i;
        }
    }

    return 0;
}

后面摆了,反正cf分还没到那个段位(\(Orz\)