AtCoder-Beginner-Contest-390

新年快乐！ 做个简单的（感觉几天没打，康复训练，新的一年，希望自己每天做题吧。 还是快乐就好

A - 12435

思路： 模拟，尝试交换每两个相邻数

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

int main() {
    int a[5] = {1, 2, 3, 4, 5};
    int b[5];
    for(int i = 0; i < 5; i++) {
        cin >> b[i];
    }
    for(int i = 0; i < 4; i++) {
        swap(b[i], b[i + 1]);
        int f = 0;
        for(int j = 0; j < 5; j++) {
            if(a[j] != b[j]) {
                f = 1;
                break;
            }
        }
        if(!f) {
            cout << "Yes\n";
            return 0;
        }
        swap(b[i], b[i + 1]);
    }
    cout << "No\n";
    return 0;
}

B - Geometric Sequence

思路：a[i] * a[i + 2] != a[i + 1] * a[i + 1] 表示等比的，小于3直接输出 Yes 。

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

signed main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    if(n < 3) {
        cout << "Yes\n";
        return 0;
    }
    for(int i = 0; i < n - 2; i++) {
        if(a[i] * a[i + 2] != a[i + 1] * a[i + 1]){
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
    return 0;
}

C - Paint to make a rectangle

思路：找到四个角的 ‘#‘， 如果这个矩形里面存在 ‘ . ’就输出No。

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

signed main() {
    int n, m;
    cin >> n >> m;
    vector<string> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    int x1 = INT_MAX, x2 = INT_MIN, y1 = INT_MAX, y2 = INT_MIN;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(a[i][j] == '#') {
                x1 = min(x1, i), x2 = max(x2, i);
                y1 = min(y1, j), y2 = max(y2, j);
            }
        }
    }
    for(int i = x1; i <= x2; i++) {
        for(int j = y1; j <= y2; j++) if(a[i][j] == '.') {
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
}

D - Stone XOR

思路：这道题就是暴搜索，也是给我教育了，也不能·太暴，搜索的时候记录三个状态，选择到哪一个数了 u， 数组记录当前分了的块 cur，used分了几块。

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

signed main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    unordered_map<int, int> mp;
    auto dfs = [&](auto&& dfs, int u, auto& cur, int used)->void {
        if(u == n) {
            int p = 0;
            for(auto x : cur) p ^= x;
            mp[p] = 1;
            return;
        }
        for(int i = 0; i < used; i++) {
            int lst = cur[i];
            cur[i] += a[u];
            dfs(dfs, u + 1, cur, used);
            cur[i] = lst;
        }
        if(used < n) {
            cur[used] = a[u];
            dfs(dfs, u + 1, cur, used + 1);
            cur[used] = 0;
        }
    };
    vector<int> cur(n);
    dfs(dfs, 0, cur, 0);
    cout << mp.size() << endl;
    return 0;
}

一开始自己写的，还是太暴力了

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

signed main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    map<int, int> mp;
    vector<int> st(n);
    // auto dfs1 = [&](auto&& dfs1, int x,)
    auto dfs = [&](auto&& dfs, int u, int t)->void {
        if(u == n) {
            mp[t]++;
            return;
        }
        int x = n - u;
        for(int i = 1; i <= x; i++) {
            auto dfs1 = [&](auto&& dfs1, int cnt, int p)->void {
                if(cnt == 0) {
                    dfs(dfs, u + i, t ^ p);
                    return;
                }
                for(int i = 0; i < n; i++) {
                    if(st[i]) continue;
                    st[i] = 1;
                    dfs1(dfs1, cnt - 1, p + a[i]);
                    st[i] = 0;
                }
            };
            dfs1(dfs1, i, 0);
        }
    };
    dfs(dfs, 0, 0);
    cout << mp.size() << endl;
    return 0;
}

E - Vitamin Balance

思路，有单调性的，可以二分，check函数背包dp找满足条件所需的价格（维生素）最小值

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

int cal(int x, auto vec) {
    vector<int> dp(m + 1);
    for(auto [b, c] : vec) {
        for(int i = m; i >= c; i--){
            dp[i] = max(dp[i], dp[i - c] + b);
        }
    }
    for(int i = 0; i <= m; i++) {
        if(dp[i] >= x) return i;
    }
    return -1;
}

signed main() {
    cin >> n >> m;
    vector<array<int, 2>> vec[3];
    for(int i = 0; i < n; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        vec[a - 1].push_back({b, c});
    }
    auto check = [&](int x)->bool {
        int res = 0;
        for(int i = 0; i < 3; i++) {
            int val = cal(x, vec[i]);
            if(val == -1) return false;
            res += val;
        }
        return res <= m;
    };
    int l = 0, r = 1e9;
    while(l < r) {
        int mid = l + r + 1 >> 1;
        if(check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l << endl;
    return 0;
}