AtCoder Beginner Contest 191

发表于2021-02-22更新于2023-09-15字数统计1.3k阅读时长14分

比赛链接

C Digital Graffiti

Description

题目链接

Solution

可以发现 (x,y)(x, y) 为顶点当且仅当 (Sx,y,Sx,y,Sx,y,Sx,y)(S_{x, y}, S_{x, y}, S_{x, y}, S_{x, y}) 中有一个或者三个为 #。于是二重循环枚举判断即可。

Code

Digital Graffiti
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#include <bits/stdc++.h>

using namespace std;
 
int main(){
    int H, W;
    cin >> H >> W;
    vector<string> S(H);
    for(auto& s : S) cin >> s;
    int ans = 0;
    for(int i = 0; i < H - 1; i++) for(int j = 0; j < W - 1; j++)
        if(S[i][j] ^ S[i][j + 1] ^ S[i + 1][j] ^ S[i + 1][j + 1]) ans++;
    cout << ans << endl;
}

D Circle Lattice Points

Description

题目链接

Solution

为防止浮点误差，将读入的小数乘以 10000 (注意不能乘 10000 后强制转化为整型)。然后枚举 xx，求出 yy 对应的取值范围，从而求出坐标为 xx 且在圆内的整数点的个数。求 y 的范围时涉及 sqrtsqrt 开根号，也会产生误差，考虑加一个偏移区间，求出 yy 的正确上下界。

也有直接求的做法 Submission;

Code

Circle Lattice Points
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#include <bits/stdc++.h>

#define debug(x) cerr << #x << " = " << x << endl

using namespace std;
typedef long long LL;

const int MAXN = 2E5 + 5;
const int MOD = 1E9 + 7;
int n, m;

LL get(LL val) {
    LL tmp = val % 10000;
    if (val >= 0)
        return val - tmp;
    else
        return val - tmp - (tmp == 0 ? 0 : 10000);
}  //返回小于等于的第一个能模10000的数

void solve(int testNum) {
    // type your code here
    string aa, bb, rr;
    LL a = 0, b = 0, r = 0;
    cin >> aa >> bb >> rr;
    int cnta = 4, cntb = 4, cntr = 4;
    bool taga = 0, tagb = 0, tagr = 0;
    for (int i = 0; i < aa.size(); i++) {
        if (taga) cnta--;
        if (aa[i] != '.')
            a = a * 10 + aa[i] - '0';
        else
            taga = 1;
    }
    while (cnta) a *= 10, cnta--;

    for (int i = 0; i < bb.size(); i++) {
        if (tagb) cntb--;
        if (bb[i] != '.')
            b = b * 10 + bb[i] - '0';
        else
            tagb = 1;
    }
    while (cntb) b *= 10, cntb--;

    for (int i = 0; i < rr.size(); i++) {
        if (tagr) cntr--;
        if (rr[i] != '.')
            r = r * 10 + rr[i] - '0';
        else
            tagr = 1;
    }
    while (cntr) r *= 10, cntr--;
    LL ans = 0;
    for (LL y = get(b - r); y <= get(b + r); y += 10000) {
        LL res = r * r - (y - b) * (y - b);
        if (res > 0) {
            LL tmp = sqrt(res);
            LL k1 = get(-tmp + a) - 100000, k2 = get(tmp + a) + 1000000;
            while ((k1 - a) * (k1 - a) > res) k1+=10000;
            while ((k2 - a) * (k2 - a) > res) k2-=10000;
            ans += (k2 - k1) / 10000 + 1;
        } else if (res == 0 && a % 10000 == 0)
            ans++;
    }
    cout << ans;
}

signed main() {
    int testCase = 1;
    for (int i = 1; i <= testCase; i++) solve(i);
    return 0;
}

E Come Back Quickly

Description

题目链接

Solution

  • 询问在有向图中从起点出发回到起点的最短回路。

  • 先求出单源最短路，由于 diststartdist_start 初始值为0，故需枚举每个点 ii，答案为 disti+wi,startdist_i + w_{i, start} 的最小值。

Code

Come Back Quickly
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#include <bits/stdc++.h>

using namespace std;

#define debug(x) cerr << #x << " = " << x << endl
#define boost ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

constexpr int MAXN = 2002, MOD = 1e9 + 7;

int n, m;
vector<pair<int, int>> g[MAXN];
array<int, MAXN> dis, inq, self;

void spfa(int st) {
    for (int i = 1; i <= n; i++) dis[i] = 1e9, inq[i] = false;
    queue<int> q;
    q.push(st), inq[st] = true;
    dis[st] = 0;
    while (!q.empty()) {
        int cur = q.front();
        q.pop();
        inq[cur] = false;
        for (auto& e : g[cur]) {
            int to = e.first, w = e.second;
            if (dis[to] > dis[cur] + w) {
                dis[to] = dis[cur] + w;
                if (!inq[to]) {
                    q.push(to);
                    inq[to] = true;
                }
            }
        }
    }
    int ans = 1e9;
    for (int i = 1; i <= n; i++) {
        for (auto & e : g[i]) {
            if (e.first == st) {
                ans = min(ans, dis[i] + e.second);
            }
        }
    }
    if (ans != 1e9) cout << ans << "\n";
    else cout << "-1\n";
}


signed main() {
    boost;
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        g[u].push_back({v, w});
    }
    for (int i = 1; i <= n; i++) spfa(i);
    return 0;
}

F GCD or MIN

Description

题目链接

Solution

  • 经过一系列操作后结果一定小于等于 min(Ai)min(A_i)。

  • 小于 min(Ai)min(A_i) 的数是通过 AA 数组中的一些数取 gcdgcd 再取最小值得到，即是一些数的约数。于是可以 O(n∗max(sqrt(Ai)))O(n * max(sqrt(A_i))) 求出小于 min(Ai)min(A_i) 的约数，然后求出其在 AA 数组中的倍数，判断它们的 gcdgcd 是否为该约数。

  • 由于 nn 个数的不同约数个数大概为 n∗log(n)n * log(n)，故总时间复杂度为 O(n∗max(sqrt(Ai))+n2∗log(n))O(n * max(sqrt(A_i)) + n^2 * log(n))。

Code

GCD or MIN
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#include <bits/stdc++.h>

using namespace std;

#define boost ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

constexpr int MAXN = 1e5 + 5, MOD = 1e9 + 7;

int n, mn;
array<int, MAXN> a;
vector<int> divs;

int main() {
    boost;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    mn = *min_element(a.begin() + 1, a.begin() + n + 1);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= sqrt(a[i]); j++) {
            if (a[i] % j == 0) divs.push_back(j), divs.push_back(a[i] / j);
        }  
    }
    sort(divs.begin(), divs.end());
    auto ed = unique(divs.begin(), divs.end());
    int ans = 0;
    for (auto it = divs.begin(); it != ed; ++it) {
        if (*it >= mn) continue;
        vector<int> vec;
        for (int i = 1; i <= n; i++) if (a[i] % (*it) == 0) vec.push_back(a[i]);
        if (vec.size() <= 1) continue;
        int gcd = vec.front();
        for (int i = 1; i < vec.size(); i++)
            gcd = __gcd(gcd, vec[i]);
        if (gcd == *it) ans++;
    }
    cout << ans + 1 << "\n";
    return 0;
}