2022.5.22

AtCoder Beginner Contest 252

A - ASCII code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
int main()
{
    int n;
    scanf("%d", &n);
    printf("%c", n);
    return 0;
}

B - Takahashi's Failure

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
int a[N], b[N];
int vis[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n;i++)
    {
        cin >> a[i];
    }
    for (int i = 1; i <= k;i++)
    {
        cin >> b[i];
        vis[a[b[i]]] = 1;
    }
    sort(a + 1, a + 1 + n);
    int f = 0;
    for (int i = n; i >= 1; i--)
    {
        if(a[i]==a[n])
        {
            if(vis[a[i]])
                f = 1;
        }
        else
            break;
    }
    if(f)
        cout << "Yes\n";
    else
        cout << "No\n";
        return 0;
}

C - Slot Strategy

枚举每个数字出现的位置以及出现相同的次数位置，对于每个数字而言最大的操作次数为j + 10 * (vis[i][j]-1))，我们需要取0-9之间的最小值。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
int vis[15][105];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n;i++)
    {
        string s;
        cin>>s;
        for (int j = 0; j < s.length();j++)
        {
            int pos = s[j] - '0';
            vis[pos][j]++;
        }
    }
    int ans = INF;
    for (int i = 0; i <= 9;i++)
    {
        int res = 0;
        for (int j = 0; j < 10; j++)
        {
            res = max(res, j + 10 * (vis[i][j]-1));
        }
        ans = min(ans, res);
    }
    cout << ans << '\n';
    return 0;
}

D - Distinct Trio

这样的三元组等价于，i<j<k,a[i]<a[j]<a[k]的个数，我们记录a[i]出现的次数，做一遍前缀和，最后累加个数。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=2e5+10,INF=1e9;
int a[N], sum[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    for (int i = 1; i <= n;i++)
    {
        cin >> a[i];
        sum[a[i]]++;
    }
    for (int i = 1;i<=2e5;i++)
    {
        sum[i] += sum[i - 1];
    }
    ll ans = 0;
    for (int i = 1; i <= n;i++)
    {
        ans += 1ll * (sum[a[i] - 1]) * 1ll * (n - sum[a[i]]);
    }
    cout << ans;
    return 0;

E - Road Reduction

跑一遍djikstra，满足d[2]+d[3]...+d[n]最小的则是遍历2-n的点，满足d[u]=d[v]+w的边v

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
struct Edge
{
    int to, w, id;
};
vector<Edge> p[N];
int dis[N];
bool vis[N];
void dijkstra()
{
    memset(vis, 0, sizeof vis);
    memset(dis, 0x3f, sizeof dis);
    priority_queue<pii, vector<pii>, greater<pii>> que;
    que.push({0,1});
    dis[1] = 0;
    while(!que.empty())
    {
        int t = que.top().second;
        que.pop();
        if(vis[t])
            continue;
        vis[t] = 1;
        for(auto tt:p[t])
        {
            int to = tt.to;
            int w = tt.w;
            if(dis[to]>dis[t]+w)
            {
                dis[to] = dis[t] + w;
                que.push({dis[to], to});
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m;
    cin >> n >> m;
    for (int i = 1; i <= m;i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        p[a].push_back({b, c, i});
        p[b].push_back({a, c, i});
    }
    dijkstra();
    for (int i = 2; i <= n;i++)
    {
        for(auto t:p[i])
        {
            int to = t.to;
            int w = t.w;
            if(dis[i]==dis[to]+w)
            {
                cout << t.id << ' ';
            }
        }
    }
        return 0;
}

F - Bread

合并果子，只是略有不同，有剩余，判断如果有剩余的话就加入小根堆中。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+10,INF=1e9;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n, l;
    cin >> n >> l;
    ll sum = 0,ans = 0;
    priority_queue<ll, vector<ll>, greater<ll>> que;
    for (int i = 1; i <= n;i++)
    {
        ll x;
        cin >> x;
        que.push(x);
        sum += x;
    }
    ll res = l - sum;
    if(res)
        que.push(res);
    while(que.size()>1)
    {
        ll a = que.top();
        que.pop();
        ll b = que.top();
        que.pop();
        ll c = a + b;
        ans += c;
        que.push(c);
    }
    cout << ans;
    return 0;
}