#include
using namespace std;
const int maxn=1e9+10;
int num[100010];
int n,m;
int judge(int dis)
{int dum=1;int xi=num[1];for(int i=2;i<=n;i++){if(num[i]-xi>=dis){dum++;xi=num[i];}if(dum==m)break;}return dum==m;}
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>n>>m;for(int i=1;i<=n;i++)cin>>num[i];sort(num+1,num+1+n);int l=num[1];int r=num[n]-num[1]+1;int k;while(l>1;if(judge(mid)){k=mid;l=mid+1;}else r=mid;}cout<
#include
using namespace std;
const int maxn=1e5+10;
int n,m;
double num1[maxn];
double num2[maxn];
bool judge(double ave)
{for(int i=1;i<=n;i++)num2[i]=num2[i-1]+num1[i]-ave;//每次都减去平均值double mins=maxn;//筛出最小前缀值for(int i=m;i<=n;i++)//从最短距离开始{mins=min(num2[i-m],mins);if(num2[i]-mins>0)return true;}return false;
}
int main()
{cin>>n>>m;double l=0,r=0;for(int i=1;i<=n;i++){cin>>num1[i];r=max(num1[i],r);}while(r-l>1e-6){double mid=(l+r)/2;if(judge(mid))l=mid;else r=mid;}printf("%d\n",(int)(1000*r));
}
#include
using namespace std;
int num;
int num1;
double a[20005],b[20005],c[20005];
double judge(double x)//返回值
{double mas=INT_MIN;for(int i=1;i<=num1;i++){mas=max(mas,a[i]*x*x+b[i]*x+c[i]);//选出最大的那个值}return mas;
}
void solves()
{cin>>num1;for(int i=1;i<=num1;i++){cin>>a[i]>>b[i]>>c[i];}double left=0,right=1000;while(right>left+1e-4)//精度{double midleft=left+(right-left)/3;//三分左区间double midright=right-(right-left)/3;//三分右区间if(judge(midleft)>=judge(midright))//如果左高的话说明左离得远{left=midleft;//移动左区间}else right=midright;//否则移动右区间}printf("%.4f\n",judge(left));}
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>num;while(num--){solves();}
}
#include
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
int n,m;
ll num[maxn];
int check(int x)
{ll sum=0;int cnt=1;for(int i=1;i<=n;i++){if(num[i]>x)return 0;sum+=num[i];if(sum>x){sum=num[i];cnt++;}}return m>=cnt;
}
int main()
{cin>>n>>m;int l=1,r=0;for(int i=1;i<=n;i++){cin>>num[i];r+=num[i];}while(l>1;if(check(mid))r=mid;else l=mid+1;}cout<
#include
using namespace std;
const int maxn=1e6+10;
int x[maxn],y[maxn];
int father[maxn];
int n;
int find(int x)
{if(father[x]==x)return x;else return find(father[x]);
}
bool check(int mid)
{int sum=0;for(int i=1;i<=n;i++)father[i]=i;for(int i=1;i<=n;i++){for(int i1=i+1;i1<=n;i1++){if(abs(x[i]-x[i1])+abs(y[i]-y[i1])<=2*mid){if(find(i)!=find(i1)){father[find(i)]=find(i1);}}}}for(int i=1;i<=n;i++){if(father[i]==i)sum++;if(sum==2)return false;}return true;}
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>n;for(int i=1;i<=n;i++){cin>>x[i]>>y[i];}int l=0,r=1e9+10;int ans;while(l>1;if(check(mid)){r=mid;}else l=mid+1;}cout<
#include
using namespace std;
double H,h,D;
double check(double x)
{if(D-x>h*x/(H-h))//影子长度小于人到墙的距离{return h*x/(H-h);//返回第一种情况}else{return D-x+h-(H-h)*(D-x)/x;//反之影子在墙上,返回第二种情况。}
}
void solve()
{cin>>H>>h>>D;double l=0;double r=D;while(r-l>1e-6){double midl=l+(r-l)/3;double midr=r-(r-l)/3;if(check(midl)>=check(midr))//单峰函数移动法{r=midr;}else l=midl;}printf("%.3f\n",check(r));
}
int main()
{int num;cin>>num;while(num--){solve();}
}
#include
using namespace std;
double Ax,Ay,Bx,By,Cx,Cy,Dx,Dy,p,q,r,ans;
double check(double x,double y)
{double P1x=Ax+(Bx-Ax)*x,P1y=Ay+(By-Ay)*x,P2x=Cx+(Dx-Cx)*y,P2y=Cy+(Dy-Cy)*y;return sqrt((P1x-Ax)*(P1x-Ax)+(P1y-Ay)*(P1y-Ay))/p+sqrt((Dx-P2x)*(Dx-P2x)+(Dy-P2y)*(Dy-P2y))/q+sqrt((P2x-P1x)*(P2x-P1x)+(P2y-P1y)*(P2y-P1y))/r;
}
double find2(double x,double l,double r)//在CD上三分
{double res=0.0;//返回的时间数while(r-l>=1e-6)//精度{double midl=l+(r-l)/3;double midr=r-(r-l)/3;double res1=check(x,midl);//按在AB中的比例,和CD中的比例返回时间double res2=check(x,midr);if(res1=1e-6){double midl=l+(r-l)/3;//在AB上三分double midr=r-(r-l)/3;double res1=find2(midl,0,1);//用三分出来的比例在CD上看返回来的值分布情况double res2=find2(midr,0,1);if(res1>Ax>>Ay>>Bx>>By>>Cx>>Cy>>Dx>>Dy>>p>>q>>r;//输入find1(0,1);//这里的0,1是在AB上的比例。printf("%.2f",ans);
}
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