Codeforces Round #409

第一题很水但是wa了一发，因为没考虑K前面是K的情况

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

#include<cstdio>

#include<cassert>

#include<iomanip>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pi acos(-1)

#define ll long long

#define mod 1000000007

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f;

bool vis[N];

int main()

{

    ios::sync_with_stdio(false);

    cin.tie();

    string s;

    cin>>s;

    int ans=;

    for(int i=;i<s.size()-;i++)

    {

        if(s[i]=='V'&&s[i+]=='K')

        {

            vis[i]=vis[i+]=;

            ans++,i++;

        }

    }

    for(int i=;i<s.size()-;i++)

        if(!vis[i]&&!vis[i+]&&s[i]=='V')

        {

            cout<<ans+<<endl;

            return ;

        }

    for(int i=;i<s.size();i++)

        if(!vis[i]&&!vis[i-]&&s[i]=='K')

        {

            cout<<ans+<<endl;

            return ;

        }

    cout<<ans<<endl;

    return ;

}

A

第二题也很水1a

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

#include<cstdio>

#include<cassert>

#include<iomanip>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pi acos(-1)

#define ll long long

#define mod 1000000007

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f;

int main()

{

    ios::sync_with_stdio(false);

    cin.tie();

    string x,z,y="";

    cin>>x>>z;

    bool f=;

    for(int i=;i<x.size();i++)

    {

        if(x[i]>=z[i])y+=z[i];

        else f=;

    }

    if(f)cout<<-<<endl;

    else cout<<y<<endl;

    return ;

}

B

第三题因为精度问题，一直wa到最后。。。。最后看别人代码时发现用k记录二分的次数，次数够大就输出-1退出，居然还有这种骚操作，唉，我居然给忘记了

话说一开始就想到二分也不错了，虽然改了很多发还是wa在第74个点

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

#include<cstdio>

#include<cassert>

#include<iomanip>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pi acos(-1)

#define ll long long

#define mod 1000000007

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f;

double a[N],b[N],p;

int n;

bool ok(double x)

{

    double s=,time=;

    for(int i=;i<=n;i++)

    {

        if(b[i]-a[i]*x<)

        {

            time+=(a[i]*x-b[i])/p;

            if(time>x)return ;

        }

    }

    return ;

}

int main()

{

    ios::sync_with_stdio(false);

    cin.tie();

    cout<<setiosflags(ios::fixed)<<setprecision();

    cin>>n>>p;

    for(int i=;i<=n;i++)cin>>a[i]>>b[i];

    double l=0.0,r=1e15+;

    int ans=;

    while(r-l>eps){

        ans++;

        if(ans>)

        {

            cout<<-<<endl;

            return ;

        }

        double m=(l+r)/;

        if(ok(m))l=m;

        else r=m;

    }

    if(l>1e15)cout<<-<<endl;

    else cout<<l<<endl;

    return ;

}

C

第四题没想到会这么水，一开始想错了，后来看了样例的解释就明白了，枚举每个点到左右两边的距离的一半的最小值就是答案了

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

#include<cstdio>

#include<cassert>

#include<iomanip>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pi acos(-1)

#define ll long long

#define mod 1000000007

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-;

const int N=+,maxn=+,inf=0x3f3f3f;

double x[N],y[N];

double dis(int a,int b,int c)

{

    double B=sqrt((y[c]-y[a])*(y[c]-y[a])+(x[c]-x[a])*(x[c]-x[a]));

    double A=(y[c]-y[a])*x[b]-(x[c]-x[a])*y[b]-(y[c]-y[a])*x[c]+(x[c]-x[a])*y[c];

    double ans=A/B;

    if(ans<)ans=-ans;

    return ans/;

}

int main()

{

    ios::sync_with_stdio(false);

    cin.tie();

    cout<<setiosflags(ios::fixed)<<setprecision();

    int n;

    cin>>n;

    for(int i=;i<=n;i++)cin>>x[i]>>y[i];

    double ans=1e18;

    for(int i=;i<=n;i++)

    {

        int x1=i-,x2=i,x3=i+;

        if(x1==)x1=n;

        if(x3==n+)x3=;

        ans=min(ans,dis(x1,x2,x3));

    }

    cout<<ans<<endl;

    return ;

}

D