2020.10.9个人赛补题

Posted 2020-12-21 yyscn

b题：b其实不难题意就是不让k个连续字符成为一个周期。这提给我的教训就是中文翻译应该结合英文题目一起看，

#include<bits/stdc++.h>
#include<algorithm>
#define ll long long
using namespace std;
const ll nl=1e5+5;
#define speed_up ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int main(){
   speed_up;
   ll n,p;
   cin>>n;
   cin>>p;
   ll i;
   string m;
   cin>>m;
   ll num=0;
   for(i=0;i<n-p;i++){
    if(m[i]!=m[i+p]||(m[i]==‘.‘&&m[i+p]==‘.‘)){
        num=1;
    }
    if(num==1){
        if(m[i]!=m[i+p]&&m[i]!=‘.‘&&m[i+p]!=‘.‘){
            break;
        }
        if(m[i]==‘.‘&&m[i+p]==‘.‘){
            m[i]=‘0‘;
            m[i+p]=‘1‘;
            break;
        }
        if(m[i]==‘.‘&&m[i+p]!=‘.‘){
            if(m[i+p]==‘1‘){
                m[i]=‘0‘;
            }else if(m[i+p]==‘0‘){
               m[i]=‘1‘;
            }
            break;
        }
        if(m[i+p]==‘.‘&&m[i]!=‘.‘){
            if(m[i]==‘1‘){
                m[i+p]=‘0‘;
            }else if(m[i]==‘0‘){
               m[i+p]=‘1‘;
            }

break;
        }
    }
   }
   if(num==0){
    cout<<"No"<<endl;
   }else{
   for(i=0;i<n;i++){
    if(m[i]==‘.‘){
        m[i]=‘0‘;
    }
    cout<<m[i];
   }
   }
   cout<<endl;
}

e题用到了数学知识，log一下简减少数字大小接即可。

#include<bits/stdc++.h>
#include<algorithm>
#define ll long long
using namespace std;
const ll nl=1e5+5;
#define speed_up ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int main(){
   speed_up;
   ll x,y;
   cin>>x;
   cin>>y;
   if(x*log(y)>y*log(x)){
    cout<<"<"<<endl;
   }else if(x*log(y)<y*log(x)){
     cout<<">"<<endl;
   }else{
   cout<<"="<<endl;
   }
}

f题：按顺序找到满足si<sj<sk,计算最小花费，直接硬做会超时，要用到dp;

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=3005;
int main(){

   int n;
    int ans=INF;
    int val[maxn],cost[maxn],f[maxn][4];
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>val[i];
    for(int i=0;i<n;i++)
        cin>>cost[i];
    memset(f,INF,sizeof(f));
    for (int i=0;i<n;i++)
    {
        f[i][0]=cost[i];
        for (int j=0;j<i;j++)
            if (val[j]<val[i])
                f[i][1]=min(f[i][1],f[j][0]+cost[i]);
        for (int j=0;j<i;j++)
            if (val[j]<val[i])
                f[i][2]=min(f[i][2],f[j][1]+cost[i]);
    }
    for(int i=0;i<n;i++)
        ans=min(ans,f[i][2]);
    if(ans!=INF)
        cout<<ans<<endl;
    else
        cout<<-1<<endl;
}