AtCoder425の感想と自分が解いたところまでの解説を書いていきます
(今回はAをpythonで書きます)
(AtCoder Beginner Contest 425)
AtCoder Beginner Contest 425

1.感想

A問題：
1回でACできて1分以内に提出できた
B問題：
Nが10までで良かった
C問題：
色々情報量が多かった

2.結果

3.解説

A問題 Sigma Cubes

C++での解法

#include <bits/stdc++.h>
using namespace std;

int main(){
    int N, ans = 0;
    cin >> N;
    
    for (int i = 1; i <= N; i++){
        ans += floor(pow(-1, i))*i*i*i;
    }
    
    cout << ans << endl;
    return 0;
}

pythonでの解法

N, ans = int(input()), 0

for i in range(1, N+1):
    ans += pow(-1, i)*i*i*i

print(ans)

B問題 Find Permutation 2

C++での解法

#include <bits/stdc++.h>
using namespace std;

int main() {
    int N;
    cin >> N;
    vector<int> A(N), P(N, -1), check;
    set<int> used;

    for (int i = 0; i < N; i++) {
        cin >> A[i];
        if (A[i] != -1) {
            P[i] = A[i];
            used.insert(A[i]);
            check.push_back(A[i]);
        }
    }

    sort(check.begin(), check.end());
    for (int i = 1; i < (int)check.size(); i++){
        if (check[i-1] == check[i]){
            puts("No");
            return 0;
        }
    }

    for (int i = 1; i <= N; i++) {
        if (used.find(i) == used.end()) {
            for (int j = 0; j < N; j++) {
                if (P[j] == -1) {
                    P[j] = i;
                    break;
                }
            }
        }
    }

    puts("Yes");
    for (int i = 0; i < N; i++) cout << P[i] << " ";
    cout << endl;

    return 0;
}

C問題 Rotate and Sum Query

C++での解法

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
    ll N, Q, offset = 0;
    cin >> N >> Q;
    vector<ll> A(N);
    for (ll &i : A) cin >> i;
    vector<ll> sum(N + 1, 0);
    for (ll i = 0; i < N; i++) sum[i+1] = sum[i]+A[i];

    for (ll _ = 0; _ < Q; _++) {
        ll type;
        cin >> type;
        if (type == 1) {
            ll c;
            cin >> c;
            offset = (offset+c)%N;
        } else {
            ll l, r, res = 0;
            cin >> l >> r;
            l--, r--;
            ll start = (l+offset)%N, end = (r+offset)%N;
            if (start <= end) res = sum[end+1]-sum[start];
            else res = (sum[N]-sum[start])+sum[end + 1];
            cout << res << endl;
        }
    }

    return 0;
}