AtCoder Beginner Contest 314

AtCoder Beginner Contest 314 - AtCoder

A - 3.14 (atcoder.jp)

题目提供了100位,所以直接用字符串输出

#include <bits/stdc++.h>
#define int long long
 
using namespace std;
 
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    string s = "3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679";
    int n;
    cin >> n;
    for(int i = 0;i < n + 2;i ++)
        cout << s[i];
 
    return 0;
}

B - Roulette (atcoder.jp)

感觉还是题意比较抽象,看懂了就能模拟出来了.

#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;
    vector<int> C[n + 1];
    for(int i = 1;i <= n;i ++){
        int x;
        cin >> x;
        for(int j = 0;j < x;j ++){
            int y;
            cin >> y;
            C[i].emplace_back(y);
        }
    }
    int X;
    cin >> X;
 
    set<pair<int,int>> s;
    for(int i = 1;i <= n;i ++){
        if(std::find(C[i].begin(), C[i].end(),X) != C[i].end()){
            s.insert({C[i].size(),i});
        }
    }
    
    if(s.size()){
        int cnt = 0;
        auto x = s.begin()->first;
        vector<int> ans;
        for(auto [i,j] : s){
            if(i == x)
                ans.emplace_back(j),cnt++;
            else
                break;
        }
        cout << cnt << '\n';
        for(auto i : ans )
            cout << i << ' ';
    }else
        cout << "0\n";
    return 0;
}

C - Rotate Colored Subsequence (atcoder.jp)

要把相同颜色都往左移一位,最后面的要移到前面来,所以我们可以预处理出相同颜色的字符串,然后最后一位放前面,我这里是逆转了一下,方便使用\(string.pop\_back()\),最后就是将对应颜色的放进去就好了

#include <bits/stdc++.h>
#define int long long
 
using namespace std;
 
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n,m;
    string s;
    cin >> n >> m >> s;
    vector<int> c(n + 1);
    s = " " + s;
    for(int i = 1;i <= n;i ++) cin >> c[i];
 
    vector<string> sc(m + 1, "");
    for(int i = 1;i <= n;i ++){
        sc[c[i]] += s[i];
    }
 
    for(auto &i : sc){
        i = i.back() + i.substr(0,i.size() - 1);
        std::reverse(i.begin(), i.end());
    }
 
    string ans = "";
    for(int i = 1;i <= n;i ++){
        ans += sc[c[i]].back();
        sc[c[i]].pop_back();
    }
    cout << ans << '\n';
    return 0;
}

D - LOWER (atcoder.jp)(离线思想)

不管前面大小写翻转了几次,只有最后一次的大小写翻转能决定之前的字符串,之后因为没有翻转了,所以单点修改即可,因为要找到最后一次大小写修改,所以我们要用到离线思想,即把每次操作存起来判断,最后再处理结果

#include <bits/stdc++.h>
#define int long long
 
using namespace std;
 
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n, m;
    string s;
    cin >> n >> s >> m;
 
    vector<int> T(m + 1),X(m + 1);
    vector<char> C(m + 1);
    for(int i = 1;i <= m;i ++)
        cin >> T[i] >> X[i] >> C[i];
 
    int f = 1;
    int pos = 0;
    for(int i = m;i >= 1;i --){
        if(T[i] != 1){
            f = T[i];
            pos = i;
            break;
        }
    }
 
    if(pos > 1){
        for(int i = 1;i < pos;i ++){
            if(T[i] == 1){
                s[--X[i]] = C[i];
            }
        }
    }
 
 
    if(f == 2)std::transform(s.begin(), s.end(),s.begin(),::tolower);
    else if(f == 3) std::transform(s.begin(), s.end(),s.begin(),::toupper);
 
    for(int i = pos + 1;i <= m;i ++){
        if(T[i] == 1){
            s[--X[i]] = C[i];
        }
    }
    cout << s << '\n';
    return 0;
}