A. Drinks Choosing

一开始卡在题意上，一开始以为是买多少，后来发现是买的能让多少人喝

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 1010;
int vis[maxn];
signed main()
{
    ios::sync_with_stdio(false);
    int n,k,tmp;
    cin >> n >> k;
    for(int i = 1;i <= n; ++i)
    {
        cin >> tmp;
        vis[tmp]++;
    }
    tmp = 0;
    for(int i = 1;i <= k; ++i)
        tmp += vis[i] / 2;
    cout << (n+1)/2-tmp+tmp*2 << endl;
    return 0;
}

B. Sport Mafia

因为放进去的个数是累加的，而且如果有糖的时候先放和先吃没有区别，找到规律就可以了

#include<bits/stdc++.h>
#define int long long
using namespace std;
signed main()
{
    ios::sync_with_stdio(false);
    int n,k,sum = 0,i;
    cin >> n >> k;
    for(i = 1;sum - (n - i + 1) < k; ++i)
        sum += i;
    cout << n - i + 1 << endl;
    return 0;
}

C. Basketball Exercise

从左到右每一列三种情况，取第一行的人，取第二行的人，和第一行第二行的人都不取。

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 1e5+10;
int a[maxn],b[maxn],dp[maxn][5];
signed main()
{
    ios::sync_with_stdio(false);
    int n,k,sum = 0,i;
    cin >> n;
    for(int i = 1;i <= n; i++)
        cin >> a[i];
    for(int i = 1;i <= n; i++)
        cin >> b[i];
    for(int i = 1;i <= n; i++)
    {
        dp[i][0] = max(max(dp[i-1][0],dp[i-1][1]),dp[i-1][2]);
        dp[i][1] = max(dp[i-1][0],dp[i-1][2]) + a[i];
        dp[i][2] = max(dp[i-1][0],dp[i-1][1]) + b[i];
    }
    cout << max(dp[n][0],max(dp[n][1],dp[n][2])) << endl;
    return 0;
}

D1. Submarine in the Rybinsk Sea (easy edition)

考虑每个数字中每个数位的贡献，然后累加起来求和

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int mod = 998244353;
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,mp,ans = 0;
    cin >> n;
    for(int i = 0;i < n; i++)
    {
        cin >> mp;
        int xx = 0,pos = 1;
        while(mp)
        {
            xx = (xx + (((mp % 10) * pos) % mod) * 11) % mod;
            pos *= 1ll*100;
            mp /= 10;
        }
        xx = (xx * n) % mod;
        ans = (ans + xx) % mod;
    }
    cout << ans << endl;
    return 0;
}

D2. Submarine in the Rybinsk Sea (hard edition)

题意和上一题是一样的，但是上一题中给出的数字都是等长的，贡献值就比较好算，这个题不等长就有点儿问题

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int mod = 998244353;
const int maxn = 1e5+10;
int cnt[15],x[maxn][15],sum;
void Cal(int tmp,int base,int y)
{
    while(tmp)
    {
        sum += base * (tmp % 10) % mod;
        sum %= mod;
        if(y > 0)
            base *= 100*1ll;
        else
            base *= 10*1ll;
        base %= mod;
        tmp /= 10;
        y--;
    }
}
signed main()
{
    //freopen("in","r",stdin);
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,xx;
    cin >> n;
    for(int i = 1;i <= n; i++)
    {
        cin >> xx;
        int tmp = 0,y = xx;
        while(y)
        {
            tmp++;
            y /= 10;
        }
        x[cnt[tmp]++][tmp] = xx;
    }
    int ans = 0;
    for(int i = 1;i <= 10; i++)
    {
        for(int j = 1;j <= 10; j++)
        {
            for(int k = 0;k < cnt[i]; k++)
            {
                if(cnt[j] == 0)
                    continue;
                sum = 0;
                Cal(x[k][i],1,j);
                Cal(x[k][i],10,j-1);
                sum  = (sum * cnt[j]) % mod;
                ans = (ans + sum) % mod;
            }
        }
    }
    cout << ans << endl;
    return 0;
}