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