AtCoder Beginner Contest 203 (A~D，玄学二分场)

补题链接：Here

A - Chinchirorin

给出 \(a,b,c\) 三个正整数，现请打印各种情况的答案：

• \(a=b=c\) ，输出一个即可
• \(a = b\ and\ a != c\) 或者 \(a = c\ and\ a != b\) 或者 \(b = c\ and\ a != b\) 输出不一样的值即可
• 三个数均不同，输出 \(0\)

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水题（RioTian是个连水题的都WA2发的FW....）

void solve() {
    int a, b, c;
    cin >> a >> b >> c;
    if (a == b) cout << c << "\n";
    else if (b == c) cout << a << "\n";
    else if (a == c) cout << b << "\n";
    else cout << "0\n";
}

B - AtCoder Condominium

包租婆有一个 \(n\) 层，每层 \(m\) 个房间的出租房，每个房间的房号：\(i0j(i\in[1,n],j\in[1,j])\)

求房号累计和

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模拟题意即可

void solve() {
    int n, k;
    cin >> n >> k;
    int sum = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= k; ++j)
            sum += (i * 100 + j);
    cout << sum ;
}

C - Friends and Travel costs

太郎决定去旅行，但他只有 \(k\) 元钱，每走过一个村庄要给 \(1\) 块钱，但太郎有 \(k\) 个朋友，如果太郎经过他朋友居住的村子的话会得到 \(B_i\) 元钱作为旅费的补充

请问太郎最远能到达多远的村子（输出村子号），

注意太郎从 \(0\) 号村子出发

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模拟，如果 \(k > a_i\) 则累加 \(b_i\)

写法上用 STL 优化

using ll = long long;
void solve() {
    ll n, k; cin >> n >> k;
    vector<pair<ll, ll>>v(n);
    for (ll i = 0; i < n; ++i)
        cin >> v[i].first >> v[i].second;
    sort(v.begin(), v.end());
    for (auto x : v) {
        if (x.first > k)break;
        k += x.second;
    }
    cout << k;
}

D - Pond

【题意待补充】

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玄学二分+玄学前缀和

const int N = 1000;
int A[800][800];
int S[801][801];
void solve() {
    int N, K;
    cin >> N >> K;
    for (int i = 0; i < N; i++)for (int j = 0; j < N; j++)cin >> A[i][j];
    int L = 0, R = 1e9 + 1;
    while (R - L > 1) {
        int mid = (L + R) / 2;
        for (int i = 0; i < N; i++)for (int j = 0; j < N; j++)S[i + 1][j + 1] = mid <= A[i][j];
        for (int i = 1; i <= N; i++)for (int j = 1; j <= N; j++) {
                S[i][j] += S[i - 1][j] + S[i][j - 1] - S[i - 1][j - 1];
            }
        bool fn = false;
        for (int i = K; i <= N; i++)for (int j = K; j <= N; j++) {
                if (S[i][j] - S[i - K][j] - S[i][j - K] + S[i - K][j - K] <= K * K / 2)fn = true;
            }
        if (fn)R = mid;
        else L = mid;
    }
    cout << L << endl;
}

E，F 由于蓝桥杯暂不填坑