传送门

A 函数

注意开long long

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define endl '\n'
const int N = 100010;
const int M = 110;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace std;

void solve() {
    int n;
    cin>>n;
    if(n&1)cout<<0<<endl;
    else cout<<(ll)pow(2,n/2)<<endl;
}


int32_t main() {
    IOS;
    int h_h;
    //cin >> h_h;
    h_h = 1;
    while (h_h--)solve();
    return 0;
}

B 病毒

感觉就是一个简单的BFS，先把感染的格子都推入队列，如何跑一边BFS即可，用一个二维数组记录每个格子被感染的先后，最后再遍历一遍输出最后被感染的格子。

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define endl '\n'
const int N = 100010;
const int M = 2010;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace std;

int dis[M][M];
int g[M][M];
int dx[] {1, 0, -1, 0};
int dy[] {0, 1, 0, -1};

void solve() {
    int n, m;
    cin >> n >> m;
    int k;
    cin >> k;
    vector<pair<int, int>> node;
    while (k--) {
        int x, y;
        cin >> x >> y;
        node.push_back({x, y});
        g[x][y] = 1;
    }
    queue<pair<int, int>> q;
    for (auto i: node) {
        q.push({i.fi, i.se});
        dis[i.fi][i.se] = 1;
    }
    while (q.size()) {
        auto t = q.front();
        q.pop();
        for (int i = 0; i < 4; i++) {
            int x = t.fi + dx[i];
            int y = t.se + dy[i];
            if (x >= 1 && x <= n && y >= 1 && y <= m && !g[x][y]) {
                dis[x][y] = dis[t.fi][t.se] + 1;
                g[x][y] = 1;
                q.push({x,y});
            }
        }
    }
    int a, b;
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            //cout << dis[i][j] << ' ';
            if (dis[i][j] > ans) {
                a = i;
                b = j;
                ans = dis[i][j];
            }
        }
        //cout << endl;
    }
//    for (int i = 1; i <= n; i++) {
//        for (int j = 1; j <= m; j++) {
//            cout << g[i][j] << ' ';
//        }
//        cout << endl;
//    }
    cout << a << ' ' << b << endl;
}


int32_t main() {
    IOS;
    int h_h;
    //cin >> h_h;
    h_h = 1;
    while (h_h--)solve();
    return 0;
}

C 鸡尾酒

贪心，我们先把每种酒按照1：1的比例算出来，以最小的那种酒作为标准，去取其他的酒，最后的结果再和V取一个min即可。

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define mpy(a,b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define endl '\n'
const int N = 100010;
const int M = 2010;
const int MOD = 1e9 + 7;
const int EPS = 1e-8;

using namespace std;

void solve() {
    int n, v;
    cin >> n >> v;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++)cin >> a[i];
    for (int i = 0; i < n; i++)cin >> b[i];
    double mn = 100010;
    for (int i = 0; i < n; i++) {
        mn = min(mn, b[i] * 1.0 / a[i]);
    }
    //cout<<mn<<endl;
    double ans = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] * mn - b[i] > EPS)ans += b[i];
        else ans += a[i] * mn;
    }
    if (ans - v >= EPS)ans = v;
    printf("%.10f\n", ans);
}


int32_t main() {
    IOS;
    int h_h;
    //cin >> h_h;
    h_h = 1;
    while (h_h--)solve();
    return 0;
}