Codeforces Round #211 (Div. 2)

ASoroban

        打表题

#include <iostream>

#include <string>
using namespace std;
const string A[10]={"O-|-OOOO","O-|O-OOO","O-|OO-OO",
                    "O-|OOO-O","O-|OOOO-","-O|-OOOO",
                    "-O|O-OOO","-O|OO-OO","-O|OOO-O",
                    "-O|OOOO-"};
int main()
{
    int n;
    cin>>n;
    if(n==0)cout<<A[0]<<endl;
    while(n>0){
        cout<<A[n%10]<<endl;
        n/=10;
    }
    return 0;
}

 

BFence

      前缀和。。。

#include <iostream>

using namespace std;
const int N=150000;
int n,k,h[N+10];
int main()
{
    cin>>n>>k;
    h[0]=0;
    int minnum=(n+1)*101,ans=1;
    for(int i=1;i<=n;i++){
        cin>>h[i];
        h[i]+=h[i-1];
        if((i>=k)&&(h[i]-h[i-k]<minnum)){
            minnum=h[i]-h[i-k];
            ans=i-k+1;
        }
    }
    cout<<ans<<endl;
    return 0;
}

CFixing Typos

        唯一的动脑子题目还让我写费事了。。。。其实可以贪的。。。。。

#include <iostream>
#include <string>

#include <cstdio>
#define  one 0
#define  two 1
using namespace std;
const int S=200000;
string s;
char A[S+10];
int  B[S+10];
int  f[S+10][2];
int main()
{
    getline(cin,s);
    int len=s.length(),p=0;
    A[0]='!';B[0]=0;
    for(int i=0;i<len;i++)
        if(s[i]==A[p]) 
            B[p]++;
        else{
            p++;
            A[p]=s[i];
            B[p]=1;
        }
    //for(int i=1;i<=p;i++)
    //    cout<<A[i]<<','<<B[i]<<endl;
    f[0][0]=f[0][1]=0;
    for(int i=1;i<=p;i++){
        f[i][one]=min(f[i-1][one],f[i-1][two])+B[i]-1;
        if(B[i]>=2)
            f[i][two]=f[i-1][one]+B[i]-2;
        else 
            f[i][two]=f[i-1][one];
    }
    //for(int i=1;i<=p;i++)cout<<B[i]<<',';cout<<endl;
    //for(int i=1;i<=p;i++)cout<<f[i][one]<<',';cout<<endl;
    //for(int i=1;i<=p;i++)cout<<f[i][two]<<',';cout<<endl;
    for(int i=p;i>=1;i--){
        if(f[i][two]<f[i][one]){
            B[i]=2;
            B[i-1]=1;
            i--;
        }
        else{
            B[i]=1;
        }
    }
    for(int i=1;i<=p;i++)
        for(int j=1;j<=B[i];j++)
            cout<<A[i];
    cout<<endl;
    return 0;
}