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牛客周赛Round 3

A

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#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
const int N=1e5+100;
#define PII pair<int,int>
bool check(string x)
{
    ll ans=0;
    rep(i,1,x.length()-1)
{
ans=ans*10+x[i]-'0';
}
return ans%7==0;
}
signed main()
{
int n;
cin>>n;
string s=to_string(n);
int m=s.length();
s=" "+s;
rep(i,0,9)
{
    string x=s;
    x+=(i+'0');
    if(check(x))
{
    rep(i,1,x.length()-1)
    {
        cout<<x[i];
    }
    return 0;
}
}
    return 0;
}

B

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//明明没做过却显示代码了（）
//直接暴力枚举26个字母即可，然后用绝对值进行计算
//注意顺时针和逆时针
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
const int N=1e5+100;
#define PII pair<int,int>
signed main()
{
    string s;
     cin>>s;
    int ans=1e9+10;
    for(int i='a';i<='z';++i){
        int res=0;
        for(int j=0;j<s.size();j++){
            int t=abs(s[j]-i);
            res+=min(t,26-t);
        }
        ans=min(res,ans);
    }
    cout<<ans<<endl;
    return 0;
}

C

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//固定一个枚举一个的思想
//傻逼了，cnt和i搞错了
//调了好一会
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
const int N=1e5+100;
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
signed main()
{
int n,m,x,y;
cin>>n>>m>>x>>y;
int cnt=min(n/2,m);
int ans=0;
rep(i,0,cnt)
{
int _cnt=min(n-i*2,(m-i)/2);
ans=max(ans,i*x+_cnt*y);
}
cout<<ans<<'\n';
    return 0;
}

D

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//贪心加等差数列直接计算即可
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define frep(i,a,n) for(int i=a;i>=n;i--)
#define int long long
const int N=1e5+100;
#define PII pair<int,int>
#define lowbit(x) (x&(-x))
const int mod=1e9+7;
signed main()
{
    int n,m;
    cin>>n;
     int ans=3*(4*n-4+1)%mod*(2*n-2)-10;
     ans%=mod;
     ans=(ans+mod)%mod;
     int num=(n*n+4-4*n)%mod;//
    ans=(ans+num%mod*2%mod*(2*n*n%mod-num+1)%mod)%mod;   
    ans=(ans+mod)%mod;
    cout<<ans<<'\n';

    return 0;
}
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