天梯赛真题补题单（L1-6 ~ L1-8）

L1-6 整除光棍(思维题)

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL N=100200,M=2020;
const double PI=3.141592;
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
        LL n;
        cin>>n;
        LL sum=0,len=0;
        while(sum<n)
        {
            sum=sum*10+1;
            len++;
        }//先计算出由111...组成的最小的大于等于n的数
        while(1)
        {
            cout<<sum/n;//我们每次从头计算除法得到一个数值
            if(sum%n==0)//求得余数看是不是0
            {
                cout<<" "<<len<<endl;//是的话直接输出长度
                break;
            }
            sum%=n;//不是0的话把余数取出来
            sum=sum*10+1;//尾巴上补一个1
            len++;
        }
    }
    return 0;
}

L1-7 装睡

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL N=100200,M=2020;
const double PI=3.141592;
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
        LL n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            string s;
            cin>>s;
            LL x,y;
            cin>>x>>y;
            if(x>=15&&x<=20&&y>=50&&y<=70) ;
            else cout<<s<<endl;
        }
    }
    return 0;
}

L1-8 矩阵A乘以B

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL N=100200,M=2020;
const double PI=3.141592;
LL a[M][M],b[M][M],c[M][M];
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
        LL an,am;
        cin>>an>>am;
        for(int i=1;i<=an;i++)
        {
            for(int j=1;j<=am;j++)
            {
                cin>>a[i][j];
            }
        }
        LL bn,bm;
        cin>>bn>>bm;
        for(int i=1;i<=bn;i++)
        {
            for(int j=1;j<=bm;j++)
            {
                cin>>b[i][j];
            }
        }
        if(am!=bn) cout<<"Error: "<<am<<" != "<<bn;
        else
        {
            cout<<an<<" "<<bm<<endl;
            for(int i=1;i<=an;i++)
            {
                for(int j=1;j<=bm;j++)
                {
                    LL ans=0;
                    for(int k=1;k<=am;k++)
                    {
                        //cout<<i<<" "<<j<<" "<<ans<<endl;
                        ans+=a[i][k]*b[k][j];
                    }
                    cout<<ans;
                    if(j!=bm) cout<<" ";
                    ans=0;
                }
                if(i!=an) cout<<endl;
            }
        }
    }
    return 0;
}