2014省赛回顾

题目链接：1058--1067

http://xcacm.hfut.edu.cn/problemset.php#

省赛题其实并不难。。。练手速用。。。

1062：

感觉是先离散化，然后从点向上下左右四个方向发射线，看和边相交的次数

1063: 4min

//4min

#include <iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
int N,t;
double ans;
double a[10010];

int main()
{
    while(cin>>N)
    {
        for (int i=0;i<N;i++)
            cin>>a[i];

        sort(a,a+N);
        //for (int i=0;i<N;i++)   cout<<a[i]<<" ";
        t=N/2-1;
        if (N%2!=0)
            ans=a[t+1];
        else
            ans=(a[t]+a[t+1])/2;

        printf("%.2lf
",ans);
    }
}

1064： 9min

//9min

#include <iostream>
#include <algorithm>
using namespace std;
int n,m,k;
int R[10000],C[10000];
int main()
{
    while(cin>>n>>m>>k)
    {
        bool ok=false;
        for (int i=1;i<=k;i++)
            cin>>R[i]>>C[i];
        for (int i=1;i<=k-1;i++)
            for (int j=i+1;j<=k;j++)
            {
                int ri=R[i],ci=C[i],rj=R[j],cj=C[j];
                int rr=abs(ri-rj),cc=abs(ci-cj);
                if((rr==1)||(cc==1))
                    ok=true;
            }
        if (ok) cout<<"YES"<<endl;  else  cout<<"NO"<<endl;
    }
    return 0;
}

1065:  24min

//24min
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

string str;
int n,q,mn;
int file[10100];

int cmps(string a,string b)
{
    int ans=0;
    for (int i=0;i<36;i++)
    {
        char aa=a[i],bb=b[i];
        if (aa!=bb)  ans++;
        if (ans>mn)    return(1000010);
    }
    return ans;
}

int main()
{
    while(cin>>n)
    {
        vector<string>  NM;
        vector<string>  HS;
        for (int i=1;i<=n;i++)
        {
            cin>>str;
            NM.push_back(str);
            cin>>str;
            HS.push_back(str);
        }
        cin>>q;
        for (int ii=1;ii<=q;ii++)
        {
            cin>>str;
            cin>>str;
            int num=0;
            mn=1000000;
            for (int i=0;i<n;i++)
            {
                string st=HS[i];
                int ans=cmps(str,st);
                //cout<<"comp:  "<<str<<" "<<st<<" , "<<ans<<endl;
                if (ans<mn)
                {
                    num=1;
                    mn=ans;
                    file[num]=i;
                }
                else if (ans==mn)
                {
                    num++;
                    file[num]=i;
                }
            }
            cout<<num<<endl;
            cout<<36-mn<<endl;
            for (int i=1;i<=num;i++)
                cout<<NM[file[i]]<<endl;
        }
    }


    return 0;
}

1066：40min

记得用那个优化：找[1,n]中满足i%p==0的i --> 枚举p的每个<=n的倍数 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
#define MAXN 100010
int np=0,n;
int prime[MAXN],phi[MAXN];
bool pr[MAXN];
long long ans;

void isprime(int n)     //求1--n的质数。pr[i]=1 : i is a prime
{
    memset(pr,true,sizeof(pr));
    int m=sqrt(n+0.5);
    pr[1]=false;
    for (int i=2; i<=m; i++)
        if (pr[i])
        {
            for (int j=i*i; j<=n; j+=i)
                pr[j]=false;
        }
    np=0;
    for (int i=1; i<=n; i++)
        if (pr[i])
        {
            np++;
            prime[np]=i;
        }
}

void calc_phi(int n)        //求1--n的欧拉函数，phi[i]=φ(i)
{
    for (int i=2; i<=n; i++)
        phi[i]=0;
    phi[1]=1;
    for (int i=2; i<=n; i++)
        if (!phi[i])
            for (int j=i; j<=n; j+=i)
            {
                if (!phi[j])    phi[j]=j;
                phi[j]=phi[j]/i*(i-1);
            }
}

int main()
{
    isprime(100000);
    calc_phi(100000);

    //cout<<np<<endl;

    while(cin>>n)
    {
        ans=0;

        for (int i=1;i<=np;i++)
        {
            int prm=prime[i];
            if (prm<=n) ans++;
            int j=2;
            while(prm*j<=n)
            {
                ans+=phi[j]*2;
                j++;
            }
        }

        cout<<ans<<endl;
    }

    return 0;
}

1067: 22min

//22min
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;

int X[10010],Y[10010];
int n,m,opt,P,Q,L;
int xadd,xmul,yadd,ymul;

int main()
{
    while(cin>>n)
    {
        for (int i=1;i<=n;i++)
            scanf("%d%d",&X[i],&Y[i]);
            //cin>>X[i]>>Y[i];
        cin>>m;
        xadd=0,xmul=1,yadd=0,ymul=1;
        for (int i=1;i<=m;i++)
        {
            scanf("%d",&opt);
            if (opt==1)
            {
                cin>>P>>Q;
                xadd+=P;
                yadd+=Q;
            }
            else if (opt==2)
            {
                cin>>L;
                xadd=xadd*L;
                yadd=yadd*L;
                xmul=xmul*L;
                ymul=ymul*L;
            }
            else if (opt==3)
            {
                ymul=ymul*-1;
                yadd=yadd*-1;
            }
            else if (opt==4)
            {
                xadd=xadd*-1;
                xmul=xmul*-1;
            }
        }
        //cout<<xadd<<" "<<xmul<<"  "<<yadd<<" "<<ymul<<endl;
        for (int i=1;i<=n;i++)
            {
                X[i]=X[i]*xmul+xadd;
                Y[i]=Y[i]*ymul+yadd;
            }
        for (int i=1;i<=n;i++)
            printf("%d %d
",X[i],Y[i]);
            //cout<<X[i]<<" "<<Y[i]<<endl;
    }
    return 0;
}