“图灵杯”

图灵杯

F题-第二大数

双指针模拟更新一遍区间范围内的ans

const int N = 1e4 + 10;
#define int long long
int a[N];

signed main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
        
    int sum = 0;
    
    for(int i  = 0;i<n;i++){
        int ans = min(a[i],a[i+1]);
        int b = max(a[i],a[i+1]);
        sum+=ans;
        for(int j = i+2;j<n;j++)
        {
            if(a[j]>ans){
                ans = a[j];
            }
            if(a[j]>b){
                ans = b;
                b = a[j];
            }
            sum+=ans;
        }
    }
    cout << sum;
}

L题-WireConnection

最小生成树板子

const int N = 2e3 + 10;
#define int long long
const int NN =  N * N;

struct node
{
    int x, y, z;
} a[N];
struct Node
{
    int x, y, w;
} s[NN];
int n;

bool cmp(struct Node a, struct Node b)
{
    return a.w < b.w;
}
int fa[N];
int find(int x)
{
    if (fa[x] != x)
        fa[x] = find(fa[x]);
    return fa[x];
}

int num = 0;
signed main()
{
    scanf("%lld", &n);
    for (int i = 0; i < n; i++)
        scanf("%lld%lld%lld", &a[i].x, &a[i].y, &a[i].z);

    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            s[num++] = {i, j, (int)ceil(sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y) + (a[i].z - a[j].z) * (a[i].z - a[j].z)))};
        }
    }
    for (int i = 0; i <= n; i++)
        fa[i] = i;
    sort(s, s + num, cmp);
    int ans = 0;
    for (int i = 0; i < num; i++)
    {
        int a = s[i].x, b = s[i].y, w = s[i].w;

        a = find(a), b = find(b);
        if (a != b)
        {
            fa[a] = b;
            ans += w;
        }
    }
    printf("%lld", ans);
    return 0;
}

J题-玄神的字符串

分情况贪心

signed main()
{
    string s;
    cin >> s;
    int n0 = 0;
    int n1 = 0;
    int n = s.size();
    for (int i = 0; i < n; i++)
    {
        if (s[i] == '0')
            n0++;
        else
            n1++;
    }
    int a, b, c;
    cin >> a >> b >> c;
    if (a <= b)
    {
        int ans = min(n0, n1) * a;
        n = (max(n0, n1) - min(n0, n1));
        if (b <= a + c)
            cout << ans + n / 2 * (b);
        else
            cout << ans + n / 2 * (a + c);
    }
    else
    {
        int ans = (n0 / 2 + n1 / 2) * b;
        if (n0 % 2 + n1 % 2)
        {
            if (a <= b + c)
                cout << ans + a;
            else
                cout << ans + (b + c);
        }
        else
            cout << ans << endl;
    }
    return 0;
}

I题-最大公约数

将所有数相加，总和的因数就是可能可以得到的答案

const int N = 5e2 + 10;
int a[N];

signed main()
{
    int n;
    scanf("%d", &n);
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        int x;
        scanf("%d", &x);
        sum += x;
    }

    int ans = 0;
    if (sum == (int)(sqrt(sum)) * (int)(sqrt(sum)))
        ans--;
    for (int i = 1; i <= sqrt(sum); i++)
        if (!(sum % i))
            ans += 2;
    cout << ans << endl;
}

K题-金牌厨师

差分+二分

const int N = 3e5 + 10;
int n, m;
struct node
{
    int x, y;
} a[N];
int vis[N];

int check(int ans)
{
    for (int i = 0; i <= n; i++)
        vis[i] = 0;
    for (int i = 0; i < m; i++)
    {
        int r = a[i].x + ans - 1;
        if (r <= a[i].y)
        {
            vis[r]++;
            vis[a[i].y + 1]--;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        vis[i] += vis[i - 1];
        if (vis[i] >= ans)
            return 1;
    }
    return 0;
}
signed main()
{
    cin >> n >> m;

    for (int i = 0; i < m; i++)
        cin >> a[i].x >> a[i].y;
    int l = 1, r = min(n, m);
    while (l < r)
    {
        int mid = (l + r + 1) >> 1;
        if (check(mid))
            l = mid;
        else
            r = mid - 1;
    }
    cout << l << endl;
}