AtCoder Beginner Contest 180 个人题解（快乐DP场）

补题链接：Here

A - box

输出

B - Various distances

按题意输出 3 种距离即可

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int N;
    ll x;
    ll ans1 = 0, ans2 = 0, ans3 = 0;
    cin >> N;
    for (int i = 0; i < N; i++) {
        cin >> x;
        ans1 += abs(x);
        ans2 += x * x;
        ans3 = max(ans3, abs(x));
    }
    double ans2_ = sqrt(double(ans2));
    cout << ans1 << endl;
    cout << fixed << setprecision(20) << ans2_ << endl;
    cout << ans3 << endl;
}

C - Cream puff

set 容器存因子，然后循环输出

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    ll x, y, a, b;
    cin >> x >> y >> a >> b;
    ll ans = 0;
    while ((a - 1) * x <= b && a * x < y && (double)a * x <= 2e18) {
        x *= a, ans++;
    }
    cout << ans + (y - 1 - x) / b << '\n';
    return 0;
}

D - Takahashi Unevolved

题意：Iroha 在游戏中有一只宠物加 Takahashi，为了帮助 Takahashi 变得强力起来，需要将它送入训练场

• Kakomon Gym：
• AtCoder Gym：

但是 STR 不能超过 Y 的情况下求 EXP 的最大值

思路：最大化去第一种 Gym 的次数然后用 Ｙ减去差值除　Ｂ　即可

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    ll x, y, a, b;
    cin >> x >> y >> a >> b;
    ll ans = 0;
    while ((a - 1) * x <= b && a * x < y && (double)a * x <= 2e18) {
        x *= a, ans++;
    }
    cout << ans + (y - 1 - x) / b << '\n';
    return 0;
}

E - Traveling Salesman among Aerial Cities

https://blog.csdn.net/weixin_45750972/article/details/109144617

题意：给定边权的计算方法，求n nn个结点的最小曼哈顿回路花费。

思路：状压DP

令#card=math&code=dp%28i%2Cj%29)为状态 下从起点出发到 的最小花费，这里的状态 指从起点要经过的城市

最后答案即为：%E2%88%921%5D%5B0%5D#card=math&code=dp%5B%281%3C%3Cn%29%E2%88%921%5D%5B0%5D)，假设起点为 。

状态方程：%5D%5Bk%5D%2Bdis(k%2Cj))%2C(i%5C%26(1%3C%3Cj)%3E0)#card=math&code=dp%5Bi%5D%5Bj%5D%3Dmin%28dp%5Bi%5D%5Bj%5D%2Cdp%5Bi%E2%88%92%281%3C%3Cj%29%5D%5Bk%5D%2Bdis%28k%2Cj%29%29%2C%28i%5C%26%281%3C%3Cj%29%3E0%29)

#include <bits/stdc++.h>
using namespace std;
using ll    = long long;
const int N = 17, M = (1 << 17) + 1;
int n, x[N], y[N], z[N];
ll dp[M][N];
ll d(int a, int b) {
    return abs(x[a] - x[b]) + abs(y[a] - y[b]) + max(0, z[b] - z[a]);
}
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> x[i] >> y[i] >> z[i];
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 1; i < (1 << n); ++i) {
        for (int j = 0; j < n; ++j)
            if ((i >> j) & 1)
                for (int k = 0; k < n; ++k)
                    dp[i][j] = min(dp[i][j], dp[i - (1 << j)][k] + d(k, j));
    }
    cout << dp[(1 << n) - 1][0] << '\n';
    return 0;
}

F - Unbranched

官方题解：https://atcoder.jp/contests/abc180/editorial/250

这里做的有点懵，贴一下代码

这里使用了 atcoder/all 头文件，添加方法：Click Here

#include <atcoder/all>
using mint = atcoder::modint1000000007;
int N, M, L;
mint f(int L) {
    mint dp[303][303];
    dp[0][0] = 1;
    for (int i = 0; i < N; i++)
        for (int j = 0; j <= i; j++) {
            mint T = dp[i][j];
            for (int k = 1; i + k <= N && j + k - 1 <= M && k <= L; k++) {
                if (k > 1) dp[i + k][j + k] += T;
                if (k == 2) T *= 500000004;
                dp[i + k][j + k - 1] += T * k;
                T *= N - i - k;
            }
        }
    return dp[N][M];
}
int main() {
    std::cin >> N >> M >> L;
    std::cout << (f(L) - f(L - 1)).val();
    return 0;
}