Codeforces Round #574 (Div. 2)补题

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A. Drinks Choosing

统计每种酒有多少人偏爱他们。 k_i 为每种酒的偏爱人数。

输出ans = (n + 1)/2 >  Σk_i / 2 ? (n + 1)/2 - Σk_i / 2 + (Σk_i / 2)  * 2 : (n + 1)/2 * 2

#include<iostream>
#include<algorithm>
using namespace std;

int n,k;
const int L = +;
int a[L];
int main()
{
    cin>>n>>k;
    int t;
    for(int i = ;i<n;++i)
    {
        cin>>t;
        a[t]++;
    }
    n =  n%? (n + )/: n/;
    int cnt = ;
    for(int i = ;i<=k;++i)
        cnt+= a[i]/;
    if(cnt<n)
        cout<< n - cnt + *cnt<<endl;
    else cout<<n*<<endl;

    return ;
}

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B. Sport Mafia

二分查找二元一次方程的解

设 放入操作数为a，取出操作数为b，有

a*(a + 1)/2 - b = k

a + b = n

联立得a*(a + 3) = 2*(n + k) 二分查找a，再解出b即可

#include<iostream>
using namespace std;

long long n,k;

int main()
{
    cin>>n>>k;
    long long lo = ;
    long long hi = 5e9;
    long long mi = (lo + hi)>>;
    while(lo <hi)
    {
        mi = (lo + hi)>>;
        if(mi * (mi + ) < *(n+k))
              lo = mi+;
        else hi = mi;
    }

        cout<<n - lo<<endl;
        return ;
    return ;
}

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C. Basketball Exercise

dp题

#include<iostream>
#include<cstring>
using namespace std;
const int L = 1e5+;
int n;
long long a[L],b[L];
long long ans;
long long dp[L][];
long long max(long long a,long long b)
{
    return a>b?a:b;
}
int main()
{
    cin>>n;
    for(int i = ;i<=n;++i)cin>>a[i];
    for(int i = ;i<=n;++i)cin>>b[i];
    dp[n][] = a[n];dp[n][] = b[n];
    dp[n-][] = dp[n][] + a[n-];
    dp[n-][] = dp[n][] + b[n-];
    for(int i = n-;i>=;--i)
    {
        dp[i][] += a[i];
        dp[i][] += b[i];
        dp[i][] += max(dp[i+][],dp[i+][]);
        dp[i][] += max(dp[i+][],dp[i+][]);
    }
    cout<<max(dp[][],dp[][]);

    return ;
}

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D1/D2. Submarine in the Rybinsk Sea

统计各串数的各个位对最终解的贡献即可。

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int L = 1e5+;
const long long MOD = ;
struct E{
    char s[];
    int l;
}str[L];
long long a[];
int n;
int NUM[];
int sum[];
long long ans;
int main()
{
    cin>>n;
    for(int i = ;i<n;++i)
    {
        cin>>str[i].s;
        str[i].l = strlen(str[i].s);
    }    

    int LEN = ;
    for(int i = ;i<n;++i)
    {
        NUM[str[i].l]++;
        LEN  = max(LEN,str[i].l);
    }
    LEN *=;
    for(int i = ;i<;++i)    sum[i] = NUM[i] + sum[i-];
    for(int i = ;i<n;++i)
    {
        reverse(str[i].s,str[i].s+str[i].l);
        for(int j = ;j<str[i].l;++j)
        {
            a[j*] += (str[i].s[j] - '')*(sum[] - sum[j]);
            a[j*]%=MOD;
            a[j* +  ] += (str[i].s[j] - '')*(sum[] - sum[j]);
            a[j*+]%=MOD;
        }
        for(int j = ;j<;++j)
        {
            if(NUM[j] == ) continue;
            int l = str[i].l - j;
            if(l<=) break;
            for(int k=;k<=l;++k)
            {
                a[j*+k-] += (str[i].s[j + k - ]-'')** (NUM[j]);
                a[j*+k-]%=MOD;
            }
        }
    }
    long long L = ;
    for(int i = ;i<LEN;++i)
    {
        ans += (L%=MOD)*(a[i]%MOD);
        ans%=MOD;
        L*=;
    }
    cout<<ans<<endl;

    return ;
}