比赛链接

A - Welcome to AtCoder Land

AC_Code

#include <bits/stdc++.h>
using namespace std;
#define int long long

signed main()
{
    string s,t;
    cin >> s >> t;
    if(s == "AtCoder" && t == "Land")
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

总结

字符串基础操作

B - Ticket Counter

AC_Code

#include <bits/stdc++.h>
using namespace std;
#define int long long
int t[110];

signed main()
{
    int n,a;
    cin >> n >> a;
    for(int i = 1; i <= n; i++)
        cin >> t[i];
    int last = 0;
    for(int i = 1; i <= n; i++)
    {
        last = max(last,t[i]) + a;
        cout << last << "\n";
    }
    return 0;
}

总结

模拟

C - Popcorn

AC_Code

#include <bits/stdc++.h>
using namespace std;
#define int long long
string s[20];
int val[20];

signed main()
{
    int n,m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
    {
        cin >> s[i];
        val[i] = 0;
        for(int j = 0; j < s[i].length(); j++)
            val[i] = val[i] * 2 + (s[i][j] == 'o' ? 1 : 0);
    }
    int ans = n;
    for(int s = 0; s < (1 << n); s++)
    {
        int now = 0,cnt = 0;
        for(int i = 1; i <= n; i++)
        {
            if(s & (1 << (i - 1)))
                now |= val[i],cnt++;
        }
        if(now == (1 << m) - 1)
            ans = min(ans,cnt);
    }
    cout << ans << "\n";
    return 0;
}

总结

位运算、状态压缩、枚举

D - Souvenirs

AC_Code

#include <bits/stdc++.h>
using namespace std;
#define int long long
int a[200010],b[200010];

signed main()
{
    int n,m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        cin >> a[i];
    for(int j = 1; j <= m; j++)
        cin >> b[j];
    sort(a + 1,a + n + 1);
    sort(b + 1,b + m + 1);
    int pos1 = 1,ans = 0;
    bool flag = true;
    for(int i = 1; i <= m; i++)
    {
        int l = pos1,r = n,tmp = -1;
        while(l <= r)
        {
            int mid = (l + r) >> 1;
            if(a[mid] < b[i])
                l = mid + 1;
            else
                r = mid - 1,tmp = mid;
        }
        if(tmp == -1 || a[tmp] < b[i])
        {
            flag = false;
            break;
        }
        ans += a[tmp];
        pos1 = tmp + 1;
    }
    if(flag)
        cout << ans << "\n";
    else
        cout << "-1\n";
    return 0;
}

总结

排序、游标、二分

E - Alphabet Tiles

AC_Code

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define mod 998244353
int c[30],dp[30][1010],C[2010][2010];

void init()
{
    for(int i = 0; i <= 2000; i++)
    {
        for(int j = 0; j <= i; j++)
        {
            if(i == j || j == 0)
                C[i][j] = 1;
            else
                C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
            C[i][j] %= mod;
        }
    }
}

signed main()
{
    int k;
    cin >> k;
    init();
    for(int i = 1; i <= 26; i++)
        cin >> c[i];
    dp[0][0] = 1;
    for(int i = 1; i <= 26; i++)
    {
        for(int j = 0; j <= k; j++)
        {
            dp[i][j] += dp[i - 1][j];
            dp[i][j] %= mod;
            for(int l = 1; l <= c[i]; l++)
            {
                if(l > j)
                    break;
                dp[i][j] += dp[i - 1][j - l] * C[j][l];
                dp[i][j] %= mod;
            }
        }
    }
    int ans = 0;
    for(int i = 1; i <= k; i++)
        ans += dp[26][i],ans %= mod;
    cout << ans << "\n";
    return 0;
}

总结

排列、动态规划、升维