KYOCERA Programming Contest 2021（AtCoder Beginner Contest 200）（AB水题，C思维，D搜索，E DP）

补题链接：Here

A - Century

整除 (200) 并且判断能否整除完全

B - 200th ABC-200

按题意即可

C - Ringo's Favorite Numbers 2

求出 (A_i - A_j) 为 200 的倍数的个数，转化为贡献值即可

using ll = long long;
void solve() {
    ll ans = 0;
    int n, a[210] = {0};
    cin >> n;
    for (ll i = 1, x; i <= n; ++i) {
        cin >> x;
        x %= 200;
        ans += a[x];
        a[x]++;
    }
    cout << ans << '
';
}

D - Happy Birthday! 2

搜索情况

int n, a[205];
vector<int> c[205], b;
void dfs(int x, int sum) {
    if (c[sum].size() && b != c[sum]) {
        cout << "Yes
";
        cout << b.size() << ' ';
        for (int i : b) cout << i << ' ';
        cout << "
";
        cout << c[sum].size() << ' ';
        for (int i : c[sum]) cout << i << ' ';
        cout << "
";
        exit(0);
    } else if (b.size())
        c[sum] = b;
    if (x == n + 1) return;
    dfs(x + 1, sum);
    b.push_back(x), dfs(x + 1, (sum + a[x]) % 200), b.pop_back();
}

void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    dfs(1, 0);
    cout << "No
";
}

E - Patisserie ABC 2

DP

using ll    = long long;
const int N = 3e6 + 10;
ll f[4][N];
void solve() {
    ll n, k;
    cin >> n >> k;
    f[0][0] = 1;
    for (int i = 1; i <= 3; i++)
        for (int j = 1; j <= 3 * n; j++)
            f[i][j] = f[i][j - 1] + f[i - 1][j - 1] - (j > n ? f[i - 1][j - 1 - n] : 0);

    ll s;
    for (s = 3; k > f[3][s]; s++)
        k -= f[3][s];

    for (int i = 2, v; i >= 1; i--) {
        for (v = 1; k > f[i][s - v]; v++)
            k -= f[i][s - v];
        cout << v << " ";
        s -= v;
    }
    cout << s << "
";
}