西南民族大学 春季 2023 训练赛 5

L1-1 自动编程

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    int n;
    cin >> n;
    printf("print(%d)\n" , n );
    return 0;
}

L1-2 太神奇了

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() {
    int a , b;
    cin >> a >> b;
    cout << a + b - 1 << "\n";
    return 0;
}

L1-3 洛希极限

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    double a, c;
    int b;
    cin >> a >> b >> c;
    if (b == 0) a = a * 2.455;
    else a = a * 1.26;
    printf("%.2lf ", a);
    if (c - a > 1e-3) cout << "^_^\n";
    else cout << "T_T\n";
    return 0;
}

L1-4 吃鱼还是吃肉

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    int n;
    cin >> n;
    while( n -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        if (a == 1) {
            if (b > 130) cout << "ni li hai! ";
            else if (b < 130) cout << "duo chi yu! ";
            else cout << "wan mei! ";

            if (c > 27) cout << "shao chi rou!\n";
            else if (c < 27) cout << "duo chi rou!\n";
            else cout << "wan mei!\n";
        } else {
            if (b > 129) cout << "ni li hai! ";
            else if (b < 129) cout << "duo chi yu! ";
            else cout << "wan mei! ";

            if (c > 25) cout << "shao chi rou!\n";
            else if (c < 25) cout << "duo chi rou!\n";
            else cout << "wan mei!\n";
        }
    }
    return 0;
}

L1-5 不变初心数

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    int n;
    cin >> n;
    while (n--) {
        int x, y = 0, z, f = 1;
        cin >> x;
        z = x * 2;
        while (z) y += z % 10, z /= 10;
        for (int i = 3, t; f && i <= 9; i++) {
            z = x * i, t = 0;
            while (z) t += z % 10, z /= 10;
            if (t != y) f = 0;
        }
        if (f) cout << y << "\n";
        else cout << "NO\n";
    }
    return 0;
}

L1-6 字母串

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    int n;
    cin >> n;
    while( n -- ){
        string s;
        cin >> s;
        bool f = true;
        for( int i = 0; f && i < s.size()-1 ; i ++ ){
            if( s[i] >= 'A' && s[i] <= 'Z' ){
                if( s[i+1] != s[i] + 'a' - 'A' && s[i+1] != s[i]+1 ) f = false;
            }else{
                if( s[i+1] != s[i] + 'A' - 'a' && s[i+1] != s[i]-1 ) f = false;
            }
        }
        if( f ) cout << "Y\n";
        else cout << "N\n";
    }
    return 0;
}

L1-7 矩阵列平移

题目的意思是\(1,2,\dots,k,1,2,\dots k,\dots\)不是\(1,k,1,k,1,k,\dots\)

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read(), k = read(), t = read();
    vector<vector<int>> a(n + 1, vector<int>(n + 1));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            a[i][j] = read();
    for (int j = 2, p = 1; j <= n; j += 2) {
        for (int i = n; i > p; i--)
            a[i][j] = a[i - p][j];
        for (int i = 1; i <= p; i++)
            a[i][j] = t;
        p++;
        if (p == k + 1) p = 1;
    }
    for (int i = 1, sum; i <= n; i++) {
        sum = 0;
        for (int j = 1; j <= n; j++) sum += a[i][j];
        cout << sum << " \n"[i == n];
    }
    return 0;
}

L1-8 均是素数

先跑个筛法,然后枚举一下素数，最后暴力判断一下素数就好了。

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

bool isPrime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (x % i) continue;
        return false;
    }
    return true;
}

int32_t main() {
    int l = read(), r = read();
    vector<bool> notPrime(r + 1);
    vector<int> pr;
    notPrime[0] = notPrime[1] = 1;
    for (int i = 2; i <= r; i++) {
        if (notPrime[i]) continue;
        for (int j = i * 2; j <= r; j += i)
            notPrime[j] = 1;
    }
    for (int i = l; i <= r; i++) {
        if (notPrime[i]) continue;
        pr.push_back(i);
    }
    int cnt = 0;
    for (int p = 0; p < pr.size(); p++)
        for (int q = p + 1; q < pr.size(); q++)
            for (int r = q + 1; r < pr.size(); r++) {
                if (isPrime(pr[p] * pr[q] + pr[r]) && isPrime(pr[q] * pr[r] + pr[p]) && isPrime(pr[r] * pr[p] + pr[q]))
                    cnt++;
            }
    cout << cnt << "\n";
    return 0;
}

L2-1 盲盒包装流水线

就是搞个小模拟

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    ios::sync_with_stdio(false),cin.tie(nullptr) , cout.tie(nullptr);
    int n , s;
    cin >> n >> s;
    vector<string> box(n);
    for( auto &i : box ) cin >> i;
    map<string,int> cnt;
    vector<int> stk(s);
    for( int t = n/s , i = 0; t ; t -- ){
        for( auto &j : stk ) cin >> j;
        reverse(stk.begin() , stk.end() );
        for( auto j : stk )
            cnt[ box[i++] ] = j;
    }
    int k;
    cin >> k;
    while( k -- ){
        string id;
        cin >> id;
        if( cnt.find(id) == cnt.end() ) cout << "Wrong Number\n";
        else cout << cnt[id] << "\n";
    }
    return 0;
}

L2-2 点赞狂魔

结构体排序

#include <bits/stdc++.h>

using namespace std;

struct Node {
    string name;
    int n;
    set<int> cnt;

    Node() {}

    bool operator<(Node b) const {
        if (cnt.size() != b.cnt.size()) return cnt.size() > b.cnt.size();
        return n < b.n;
    }
};

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n;
    cin >> n;
    vector<Node> a(n);
    for (auto &i: a) {
        cin >> i.name >> i.n;
        for (int j = 1, x; j <= i.n; j++)
            cin >> x, i.cnt.insert(x);
    }
    sort(a.begin(), a.end());
    for (int i = 0; i < min(3, n); i++)
        cout << a[i].name << " \n"[i == 2];
    for (int i = max(0, 3 - n); i; i--)
        cout << "-" << " \n"[i == 1];
    return 0;
}

L2-3 浪漫侧影

把二叉树建好，然后跑一下层序遍历

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

struct Node {
    int v;
    Node *l, *r;

    Node(int v, Node *l, Node *r) : v(v), l(l), r(r) {};
};

Node *build(vector<int> mid, vector<int> suf) {
    int v = suf.back();
    Node *l = nullptr, *r = nullptr;
    int t;
    for (t = 0; t < mid.size(); t++)
        if (mid[t] == v) break;
    auto midll = mid.begin();
    auto midlr = mid.begin() + t;
    auto midrl = mid.begin() + t + 1;
    auto midrr = mid.end();
    auto sufll = suf.begin();
    auto suflr = suf.begin() + t;
    auto sufrl = suf.begin() + t;
    auto sufrr = suf.end() - 1;

    auto midl = vector<int>(midll, midlr);
    auto midr = vector<int>(midrl, midrr);
    auto sufl = vector<int>(sufll, suflr);
    auto sufr = vector<int>(sufrl, sufrr);
    
    if (!midl.empty()) l = build(midl, sufl);
    if (!midr.empty()) r = build(midr, sufr);

    return new Node(v, l, r);
}


int32_t main() {
    int n = read();
    vector<int> mid(n), suf(n);
    for (auto &i: mid) i = read();
    for (auto &i: suf) i = read();
    auto v = vector<int>(mid.begin(), mid.begin() + 3);
    Node *root = build(mid, suf);
    vector<int> L, R;
    map<int, int> dep;

    queue<pair<Node *, int>> q;
    dep[root->v] = 1, q.emplace(root, 1);
    while (!q.empty()) {
        auto [u, d] = q.front();
        q.pop();
        if (L.empty()) L.push_back(u->v);
        else if (d != dep[L.back()]) L.push_back(u->v);
        if (R.empty()) R.push_back(u->v);
        else if (d != dep[R.back()]) R.push_back(u->v);
        else R.back() = u->v;
        if (u->l != nullptr) {
            dep[u->l->v] = d + 1;
            q.emplace(u->l, d + 1);
        }
        if (u->r != nullptr) {
            dep[u->r->v] = d + 1;
            q.emplace(u->r, d + 1);
        }
    }
    cout << "R:";
    for (auto i: R) cout << " " << i;
    cout << "\nL:";
    for (auto i: L) cout << " " << i;
    return 0;
}

L2-4 哲哲打游戏

把图建出来然后模拟这个过程就好了

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read() , m = read();
    vector<vector<int>> e(n+1);
    for( int i = 1 , k ; i <= n ; i ++ ){
        k = read();
        e[i] = vector<int>(k+1);
        for( int j = 1 ; j <= k ; j ++ ) e[i][j] = read();
    }
    vector<int> s(101,-1);
    int t = 1;
    for( int op , x ; m ; m -- ){
        op = read() , x = read();
        if( op == 0 ) t = e[t][x];
        else if( op == 1 )
            s[x] = t , printf("%d\n" , t);
        else t = s[x];
    }
    printf("%d\n" , t );
    return 0;
}

L3-1 直捣黄龙

首先从终点跑一遍 dij，然后从七点开始跑一遍暴搜就可。

#include <bits/stdc++.h>

using namespace std;

const int N = 205;
int n, k, s, e;
string sta, ed;
int val[N], dis[N], vis[N];
vector<pair<int, int>> g[N];
unordered_map<string, int> sti;
unordered_map<int, string> its;

vector<int> ans, now;
int ansdis = -1, anscnt = -1 , anstimes;


void dij() {
    fill(dis + 1, dis + 1 + n, INT_MAX);
    dis[e] = 0;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
    q.emplace(0, e);
    while (!q.empty()) {
        auto [d, u] = q.top();
        q.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        for (auto [v, w]: g[u]) {
            if (vis[v]) continue;
            if (dis[v] <= d + w) continue;
            dis[v] = d + w, q.emplace(dis[v], v);
        }
    }
    return;
}

void dfs(int x, int d, int cnt) {
    if (d + dis[x] > ansdis) return;
    if (x == e && d == ansdis ) {
        anstimes ++;
        if (now.size() > ans.size()) {
            anscnt = cnt, ans = now;
        } else if (now.size() == ans.size() && cnt > anscnt) {
            anscnt = cnt, ans = now;
        }
        return;
    }
    for (auto [v, w]: g[x]) {
        if (vis[v]) continue;
        now.push_back(v), vis[v] = 1;
        dfs(v, d + w, cnt + val[v]);
        now.pop_back(), vis[v] = 0;
    }
    return;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> k >> sta >> ed;
    sti[sta] = 1, its[1] = sta;
    for (int i = 2; i <= n; i++)
        cin >> its[i] >> val[i], sti[its[i]] = i;
    s = sti[sta], e = sti[ed];
    for (int u, v, w; k; k--) {
        string us, vs;
        cin >> us >> vs >> w;
        u = sti[us], v = sti[vs];
        g[u].emplace_back(v, w);
        g[v].emplace_back(u, w);
    }

    dij();
    fill(vis, vis + 1 + n, 0);
    ansdis = dis[s], vis[s] = 1;
    dfs(s, 0, 0);
    cout << sta;
    for (auto i : ans)
        cout << "->" << its[i];
    cout << "\n" << anstimes << " " << ansdis << " " << anscnt << "\n";
    return 0;
}

L3-2 拼题A打卡奖励

正常的想法是就是01背包，\(f[i]\)表示用时\(i\)能否获得最多的金币，但是\(i\)的取值太大了，这里可以改成表示为获得金币\(i\)的最少用时，倒序遍历第一个小于\(f[i]<m\)的\(i\)就是答案

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read(), m = read(), k = 0;
    vector<int> w(n), v(n);
    for (int &i: w) i = read();
    for (int &i: v) i = read(), k += i;
    vector<int> f(k + 1, INT_MAX);
    f[0] = 0;
    for (int i = 0; i < n; i++) {
        for (int j = k; j >= v[i]; j--) {
            if (f[j - v[i]] == INT_MAX) continue;
            f[j] = min(f[j], f[j - v[i]] + w[i]);
        }
    }

    for (int i = k; i >= 0; i--)
        if (f[i] <= m) cout << i, exit(0);
    return 0;
}