天梯赛L2题解（053-056）

L2-053 算式拆解

如果你WA2了，注意数字可能很长，超出longlong范围
但是实际上我们并不需要算出来的数字，所以用字符串存就好了
甚至有更简单好写的做法，但我懒得写了

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

void ylh_() {
    string s;
    cin >> s;
    int n = s.length();
    stack<string> num;
    stack<char> op;
    auto work = [&]() -> void {
        auto b = num.top();
        num.pop();
        auto a = num.top();
        num.pop();
        char c = op.top();
        op.pop();
        int res = 0;
        if (b == "OooOOOO00O0o" && a == "OooOOOO00O0o") {
            cout << c << '\n';
        } else if (a == "OooOOOO00O0o" && b != "OooOOOO00O0o") {
            cout << c << b << '\n';
        } else if (b == "OooOOOO00O0o" && a != "OooOOOO00O0o") {
            cout << a << c << '\n';
        } else {
            cout << a << c << b << '\n';
        }
        num.push({ "OooOOOO00O0o" });
    };

    for (int i = 0; i < n; ++i) {
        if (isdigit(s[i])) {
            int j = i;
            string cur;
            while (j < n && isdigit(s[j])) {
                cur += s[j];
                ++j;
            }
            i = j - 1;
            num.push({ cur });
        } else if (s[i] == '(') {
            continue;
        } else if (s[i] == ')') {
            work();
        } else {
            op.push(s[i]);
        }
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-054 三点共线

如果只枚举两个点那么复杂度会大大减少，因为题目按照1,0排序，所以我们按这个顺序枚举就行，卡卡常剪剪枝就能过

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

const int N = 1e6;
int st[3][N * 2 + 10];//全局数组更快

void ylh_() {
    int n;
    cin >> n;
    vector<vector<int>> seg(3);
    for (int i = 1; i <= n; ++i) {
        int x, y;
        cin >> x >> y;
        if (!st[y][x + N]) {
            seg[y].push_back(x);
        }
        st[y][x + N] = 1;
    }
    for (int i = 0; i <= 1; ++i) {
        sort(seg[i].begin(), seg[i].end());
    }
    auto print = [&](int x1, int y1, int x2, int y2, int x3, int y3) {
        cout << '[' << x1 << ", " << y1 << "] ";
        cout << '[' << x2 << ", " << y2 << "] ";
        cout << '[' << x3 << ", " << y3 << "]\n";
    };
    int f = 0;
    for (int x1 : seg[1]) {
        for (int x0 : seg[0]) {//存答案还要排序可能更慢,这样不用存
            int x2 = 2 * x1 - x0;
            if (x2 > N)
                continue;
            if (x2 < -N)
                break;//自己手画一下发现这个可以改成break，是一个剪枝
            if (st[2][x2 + N]) {
                f = 1;
                print(x0, 0, x1, 1, x2, 2);
            }
        }
    }
    if (!f) {
        cout << -1;
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-055 胖达的山头

简单差分

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

void ylh_() {
    int n;
    cin >> n;
    vector<int> a(24 * 3600 + 10);
    for (int i = 1; i <= n; ++ i) {
        int h, m, s, time;
        char c;
        cin >> h >> c >> m >> c >> s;
        time = h * 3600 + m * 60 + s;
        a[time]++;
        cin >> h >> c >> m >> c >> s;
        time = h * 3600 + m * 60 + s;
        a[time + 1]--;
    }
    int ans = a[0];
    for (int i = 1; i <= 24 * 3600; ++ i) {
        a[i] += a[i - 1];
        ans = max(a[i], ans);
    }
    cout << ans;
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-056 被n整除的n位数

感觉答案不会太多，所以爆搜试试，发现可过

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

void ylh_() {
    int n, a, b;
    cin >> n >> a >> b;
    int f = 0;
    int val = 0;
    auto dfs = [&](auto&& dfs, int w) {
        if (w == n) {
            if (val >= a && val <= b) {
                f = 1;
                cout << val << '\n';
            }
            return;
        }
        for (int i = 0; i <= 9; ++i) {
            if (w == 0 && i == 0)
                continue;
            val = val * 10 + i;
            if (val % (w + 1) == 0) {
                dfs(dfs, w + 1);
            }
            val /= 10;
        }
    };
    dfs(dfs, 0);
    if (!f) {
        cout << "No Solution";
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2总算完结了