Jan.17th
注：黑粗体为超链接
一道纯物理题……算得蛋疼
C++ Code
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#include<cstdio>
#include<cstring> #include<algorithm> #include<cctype> #include<ctime> #include<cstdlib> #include<string> #include<queue> #include<cmath> #define Rep(x,a,b) for (int x=a;x<=b;x++) #define Per(x,a,b) for (int x=a;x>=b;x--) #define ll long long using namespace std; inline int IN(){ int x=0,ch=getchar(),f=1; while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar(); if (ch=='-'){f=-1;ch=getchar();} while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } double a,v,l,d,w; int main(){ a=1.0*IN(),v=1.0*IN(),l=1.0*IN(),d=1.0*IN(),w=1.0*IN(); if (w>=v||2*a*d<=w*w){ if (v*v>=2*a*l) return printf("%.15lf\n",sqrt(2*a*l)/a),0; else return printf("%.15lf\n",(l-v*v/(2*a))/v+v/a),0; } else{ double t=0,pred=v*v/(2*a)+(v*v-w*w)/(2*a); if (pred<=d) t+=v/a+(v-w)/a+(d-pred)/v; else{ double v0=sqrt(a*d+w*w/2); t+=v0/a+(v0-w)/a; } double sufd=(v*v-w*w)/(2*a); if (sufd<=l-d){ t+=(l-d-sufd)/v+(v-w)/a; } else{ double v0=sqrt(2*a*(l-d)+w*w); t+=(v0-w)/a; } printf("%.15lf\n",t); } } |
这题都不会来……人生了无希望
先把这个环从最高的那个山丘上断开变成一条链 这是等价的
然后l数组记录每个点在它左边离它最近的比它高的山丘的编号
r数组记录每个点在它右边离它最近的比他高的山丘的编号
ci记录i到ri之间高度和它一样的山的数量
然后对于每座山丘ci先加到Ans中去
然后左边右边判一下加一加
具体见代码
C++ Code
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#include<cstdio>
#include<cstring> #include<algorithm> #include<cctype> #include<ctime> #include<cstdlib> #include<string> #include<queue> #include<cmath> #define Rep(x,a,b) for (int x=a;x<=b;x++) #define Per(x,a,b) for (int x=a;x>=b;x--) #define ll long long using namespace std; inline int IN(){ int x=0,ch=getchar(),f=1; while (!isdigit(ch)&&(ch!='-')&&(ch!=EOF)) ch=getchar(); if (ch=='-'){f=-1;ch=getchar();} while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } const int maxn=1e6+5; int n,Max,pos; int a[maxn],b[maxn],l[maxn],r[maxn],c[maxn]; int main(){ n=IN();Max=0;pos=0; Rep(i,1,n){ a[i]=IN(); if (a[i]>=Max){Max=a[i];pos=i;} } Rep(i,pos,n) b[i-pos+1]=a[i]; Rep(i,1,pos-1) b[i+1+n-pos]=a[i]; b[n+1]=a[pos]; l[1]=0; Rep(i,2,n){ l[i]=i-1; while (l[i]>1&&b[l[i]]<b[i]) l[i]=l[l[i]]; if (b[i]==b[l[i]]) l[i]=l[l[i]]; } c[n+1]=0; Per(i,n,1){ r[i]=i+1; while (r[i]<n+1&&b[r[i]]<b[i]) r[i]=r[r[i]]; if (b[r[i]]==b[i]){c[i]=c[r[i]]+1;r[i]=r[r[i]];} } ll Ans=0; Rep(i,2,n){ Ans+=c[i]; if (b[i]!=Max){ if (l[i]==1&&r[i]==n+1) Ans++; else Ans+=2; } } printf("%I64d\n",Ans); } |
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