2020 Smart Calculation Road Preliminary Round 2

A. Sound control light

#pragma GCC optimize(2)
#include <cstdio>

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. Construct a string

#pragma GCC optimize(2)
#include <cstdio>

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. Intelligence War

Idea: Put the two numbers of the known sum into the same set through the union search set, and then open two arrays to record whether each set is accessed and the elements in the set. Note that when x == y, it can be regarded as case 1.

#pragma GCC optimize(2)
#include <cstdio>
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;
}