The 6th Zhejiang Provincial Collegiate Programming Contest

Second-price Auction ：稳定排序水过。

#include<iostream> #include<cstdio> #include<algorithm> #include<cstdio> using namespace std; class node { public: int no,date; }; node tree[105]; bool cmp(node a,node b) { if(a.date >b.date )return true; return false; } int main() { int t,n,i; cin>>t; while(t--) { cin>>n; for(i=1;i<=n;i++) { scanf("%d",&tree[i].date ); tree[i].no =i; } sort(tree+1,tree+1+n,cmp); printf("%d %d\n",tree[1].no ,tree[2].date ); } return 0; }

Light Bulb：三分法。注意精度问题。

#include<iostream> #include<cstdio> #include<string> using namespace std; #define esp 1e-6 double H,h,D; double suc(double x) { return (h*D-H*x)/(D-x)+x; } int main() { int t; cin>>t; while(t--) { cin>>H>>h>>D; double left=0,right=D*h/H; while(right-esp>left) { double mid=(left+right)/2.0; double midmid=(mid+right)/2.0; double c1=suc(mid); double c2=suc(midmid); if(c1>=c2) right=midmid; else left=mid; } printf("%.3lf\n",suc(left)); } return 0; }

80ers' Memory：这种题，很适合用set。

#include<iostream> #include<string> #include<cstdio> #include<set> using namespace std; set<string>tree; int main() { int flag,n,i,k; string str; cin>>n; for(i=0;i<n;i++) { cin>>str; tree.insert (str); } cin>>k; while(k--) { cin>>n; flag=0; while(n--) { cin>>str; if(tree.find (str)!=tree.end ()) flag++; } printf("%d\n",flag); } return 0; }

A Stack or A Queue?：模拟题。

#include<iostream> #include<cstdio> #include<string> #include<stack> #include<queue> #include<vector> using namespace std; int a1[105],a2[105]; int main() { int t,i,n,k; cin>>t; while(t--) { queue<int> q; stack<int> st; cin>>n; for(i=1;i<=n;i++) { scanf("%d",&a1[i]); q.push (a1[i]); st.push (a1[i]); } for(i=1;i<=n;i++) scanf("%d",&a2[i]); i=1; while(!q.empty ()) { if(a2[i]==q.front ()) { i++; q.pop (); } else break; } i=1; while(!st.empty ()) { if(a2[i]==st.top ()) { i++;st.pop ();} else break; } if(q.empty ()&&st.empty ()) cout<<"both"<<endl; else if(q.empty ()&&!st.empty ()) cout<<"queue"<<endl; else if(!q.empty ()&&st.empty ()) cout<<"stack"<<endl; else cout<<"neither"<<endl; } return 0; }

K-Nice：这题比较有意思，在梦中我感觉好像看了解题报告，醒来之后，真的想到了对策！

#include<iostream> #include<cstdio> #include<string> using namespace std; int a[16][16]; int dir[][2]={{-1,0},{0,-1},{0,1}}; int f(int tx,int ty) { int sum=a[tx][ty]; for(int i=0;i<3;i++) { int xx=tx+dir[i][0]; int yy=ty+dir[i][1]; sum-=a[xx][yy]; } return sum; } int main() { int t,i,j; int n,m,k; cin>>t; while(t--) { cin>>n>>m>>k; if(k==(n-2)*(m-2)) { for(i=1;i<=n;i++) { for(j=1;j<m;j++) printf("%d ",0); printf("%d\n",0); } continue; } for(i=1;i<=m;i++) { a[1][i]=0;a[2][i]=0; } for(i=1;i<=n;i++) { a[i][1]=0;a[i][m]=0; } int num=0; for(i=3;i<=n;i++) { for(j=2;j<m;j++) { if(k) { k--; a[i][j]=f(i-1,j); } else { int temp=f(i-1,j); int aa=1; while(aa==temp) aa++; a[i][j]=aa; } } } for(i=1;i<=n;i++) { for(j=1;j<m;j++) printf("%d ",a[i][j]); printf("%d\n",a[i][m]); } } return 0; }