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Niu Ke Xiaobaiyue Race 53 A-E

2022-07-30 21:06:00 【Salted egg_dd】

A-Raining

B-Kissing

A-Raining

原题链接<<See the original title here

思路：简单略

#include <bits/stdc++.h>
using namespace std;
int main()
{
   int a,b,c,d,x,ans;
   ans=0;
   scanf("%d%d%d%d",&a,&b,&c,&d);
   scanf("%d",&x);
   if(a<x)
       printf("%d ",x-a);
    else
        printf("0 ");
   if(b<x)
       printf("%d ",x-b);
    else
        printf("0 ");
  if(c<x)
       printf("%d ",x-c);
    else
        printf("0 ");
   if(d<x)
       printf("%d ",x-d);
    else
        printf("0 ");
   return 0;
}

B-Kissing

原题链接-See the original title here

思路：Expand the formula and simplify it,根据1-nThe summation formula can be found beforen项为(n)*(n+1)-n,Because the answer will be very large, it is necessary to model it,But the process volume will be huge,So the final result model is not acceptable,Every item must be modeled

#include <bits/stdc++.h>
using namespace std;
const int N=998244353;
int main()
{
    long long int n;
    scanf("%lld",&n);
    printf("%lld\n",((n%N)*((n+1)%N)-n%N)%N);

return 0;
}

C. Missing

See the original title here

思路：模拟题,Just write as the title says

注意 The lexicographical output is availablestring,因为可以直接sort

#include <bits/stdc++.h>
using namespace std;
const int N=110;
string s,ss[N],sss[N];
queue <int> q;
int main()
{
    int n,i;
    cin>>s;
    cin>>n;
    int len=s.length();
    for(i=1;i<=n;i++)
    {
        cin>>ss[i];
    }
    int res=0;
    for(i=1;i<=n;i++)
    {
        int k=ss[i].length();
        if(k!=len)
            continue;
        int sum=0;
        for(int j=0;j<len;j++)
        {
            if(ss[i][j]==s[j])
                sum++;
        }
        if(res<sum)
        {
            res=sum;
            while(!q.empty())
                q.pop();
            q.push(i);
        }
        else if(res==sum)
            q.push(i);
    }
    if(res==0)
    {
        sort(ss+1,ss+1+n);
        for(i=1;i<=n;i++)
            cout<<ss[i]<<endl;
    }
    else
    {
        int sum=q.size();
        i=1;
        while(!q.empty())
        {
            sss[i]=ss[q.front()];
            q.pop();
            i++;
        }
        sort(sss+1,sss+1+sum);
        for(i=1;i<=sum;i++)
            cout<<sss[i]<<endl;
    }
    return 0;
}

D.Breezing

思路：The difference between each two numbers is the largest,Then just take it1或者bi,运用dp,The two cases are counted separately,Take the maximum value at the end

#include <bits/stdc++.h>
using namespace std;
const int N=100010;
typedef long long int ll;
int main()
{
    ll n,i;
    ll a[N],dp[N][3];//dp[i][1]表示第i个数取1的答案
    cin>>n;          //dp[i][2]表示第i个数取b[i]的答案
    for(i=1;i<=n;i++) cin>>a[i];
    dp[1][1]=0;
    dp[1][2]=0;
    for(i=2;i<=n;i++)
    {
        dp[i][1]=max(dp[i-1][1],dp[i-1][2]+a[i-1]-1);
        dp[i][2]=max(dp[i-1][1]+a[i]-1,dp[i-1][2]+abs(a[i]-a[i-1]));
    }
    cout<<max(dp[n][1],dp[n][2])<<endl;
    return 0;
}

E.calling

思路：贪心思想,我们会发现,When the side length of the square is 6的时候,没有空余,边长是5的时候,会剩下11A cube that becomes one,边长为4时,会剩下5one becomes2的方块,以此类推,We fill each square,就是最好的情况.

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
int main()
{
    ll t,k,n,s,i;
    ll a[10];
    cin>>t;
    while(t--)
    {
        cin>>s;
        for(i=1;i<=6;i++) cin>>a[i];
        if(a[6]!=0)
            s=s-a[6];
        if(a[5]!=0)
        {
            s=s-a[5];
            a[1]=a[1]-a[5]*11;
        }
        if(a[4]!=0)
        {
            s=s-a[4];
                 a[2]=a[2]-5*a[4];
                if(a[2]<=0)
                {
                    a[1]-=-a[2]*4;
                    a[2]=0;
                }
        }
        if(a[3]!=0)
        {
            if(a[3]%4==0)
                s=s-a[3]/4;
            else
            {
                s=s-a[3]/4-1;
                k=a[3]%4;
                if(k==1)
                {
                    if(a[2]>0)
                    {
                        a[2]=a[2]-5;
                        a[1]=a[1]-7;
                    }

                    else a[1]-=27;
                    if(a[2]<0)
                    {
                        a[1]-=-a[2]*4;
                        a[2]=0;
                    }

                }
                else if(k==2)
                {
                    if(a[2]>0)
                    {
                        a[2]=a[2]-3;
                        a[1]=a[1]-6;
                    }
                    else a[1]-=18;
                    if(a[2]<0)
                    {
                        a[1]-=-a[2]*4;
                        a[2]=0;
                    }

                }
                else if(k==3)
                {
                    if(a[2]>0)
                    {
                        a[2]=a[2]-1;
                        a[1]=a[1]-5;
                    }

                    else a[1]-=9;
                    if(a[2]<0)
                    {
                        a[1]-=-a[2]*4;
                        a[2]=0;
                    }

                }
            }
        }
        if(a[2]>0)
        {
            if(a[2]%9==0)
                s=s-a[2]/9;
            else
            {
                s=s-a[2]/9-1;
                k=a[2]%9;
                a[1]=a[1]-(9-k)*4;
            }
        }
        if(a[1]>0)
        {
            if(a[1]%36==0)
            s=s-a[1]/36;
            else
                s=s-a[1]/36-1;
        }
        //for(i=1;i<=6;i++)
            //cout<<a[i]<<' ';
        if(s>=0)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;

    }
    return 0;
}