2020智算之道初赛第二场

A.声控灯

#pragma GCC optimize(2)
#include 

using namespace std;

int main()
{
    int t;
    scanf("%d", &t); 
    
    int n, m, nums[5];
    for (int i = 0; i < t; i++)
    {
        scanf("%d%d", &n, &m);
        for (int j = 0; j < m; j++)
            scanf("%d", &nums[j]);
        
        if (m == 3) printf("%d\n", nums[1]);
        else if (m == 1) printf("1\n");
        else if (m == 2 && n != 2 && nums[0] == 1) printf("1\n");
        else if (m == 2 && n != 2 && nums[1] == n) printf("%d\n", n);
        else printf("-1\n");
    } 
    
    return 0;
}

b.构造字符串

#pragma GCC optimize(2)
#include 

using namespace std;
typedef long long LL;
LL A[26];

int main()
{
    LL n;
    scanf("%lld", &n);

    for (int i = 0; i < 26; i++)
        scanf("%d", &A[i]);

    LL x = 0;
    for (int i = 0; i < 26; i++)
        x += A[i]/n;

    printf("%lld", x);
    return 0;
}

c.情报战

思路：通过并查集将已知和的两数放入同一集合，再开两个数组分别记录各个集合是否被访问及集合内的元素。注意，当x==y时可视为情况1。

#pragma GCC optimize(2)
#include 
using namespace std;

const int N = 300010;

int f[N], vis[N], cnt[N];

inline int find (int x)
{
    if (x != f[x]) x = find(f[x]);
    return f[x];
}

int main()
{

    int n, m;
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i++) f[i] = i, cnt[i] = 1;

    int op, x, y, ans = 0;
    for (int i = 0; i < m; i++)
    {
        scanf("%d", &op);

        if (op == 1)
        {
            scanf("%d", &x);
            int fx = find(x);
            if (!vis[fx]) ans += cnt[fx], vis[fx] = 1;
        }
        else if (op == 2)
        {
            scanf("%d%d", &x, &y);
            if(x == y)
            {
                int fx = find(x);
                if (!vis[fx]) ans += cnt[fx], vis[fx] = 1;
            }
            else
            {
                int fx = find(x), fy = find(y);
                if(fx != fy)
                {
                    if (!vis[fx] && vis[fy]) ans += cnt[fx];
                    else if(vis[fx] && !vis[fy]) ans += cnt[fy], vis[fy] = 1;

                    cnt[fy] += cnt[fx];
                    f[fx] = fy;
                }
            }
        }

        printf("%d\n", ans);
    }

    return 0;
}