Codeforces Beta Round #12 (Div 2 Only)

Codeforces Beta Round #12 (Div 2 Only)

http://codeforces.com/contest/12

A

水题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 string str[];

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     for(int i=;i<;i++){
         cin>>str[i];
     }
     for(int i=;i<;i++){
         for(int j=;j<;j++){
             if(str[i][j]!=str[-i][-j]){
                 cout<<"NO"<<endl;
                 return ;
             }
         }
     }
     cout<<"YES"<<endl;

     return ;
 }

B

水题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 string str[];
 int ch[];

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("1.txt","r",stdin);
     #endif
     string str1,str2,ans="";
     cin>>str1>>str2;
     for(int i=;i<str1.length();i++){
         ch[str1[i]-'']++;
     }
     for(int i=;i<;i++){
         if(ch[i]){
             ans+=char(i+'');
             ch[i]--;
             break;
         }
     }

     for(int i=;i<;i++){
         for(int j=;j<ch[i];j++){
             ans+=char(i+'');
         }
     }
    // cout<<ans<<" "<<str2<<endl;
     if(ans==str2) cout<<"OK"<<endl;
     else cout<<"WRONG_ANSWER"<<endl;
     return ;
 }

C

水题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int n,m;
 map<string,int>mp;

 int a[];
 bool cmp1(int a,int b){
     return a>b;
 }

 bool cmp2(int a,int b){
     return a<b;
 }

 bool cmp(pair<int,string>a,pair<int,string>b){
     return a.first>b.first;
 }

 vector<pair<int,string> >ve;

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     cin>>n>>m;
     string str;
     for(int i=;i<n;i++) cin>>a[i];
     for(int i=;i<m;i++){
         cin>>str;
         mp[str]++;
     }
     map<string,int>::iterator it;
     for(it=mp.begin();it!=mp.end();it++){
         ve.push_back(make_pair(it->second,it->first));
     }
     sort(a,a+n,cmp2);
     int ans1=,ans2=;
     sort(ve.begin(),ve.end(),cmp);
    /* for(int i=0;i<ve.size();i++){
         cout<<ve[i].first<<" "<<ve[i].second<<endl;
     }*/
     for(int i=;i<ve.size();i++){
         ans1+=ve[i].first*a[i];
     }
     sort(a,a+n,cmp1);
     for(int i=;i<ve.size();i++){
         ans2+=ve[i].first*a[i];
     }
     cout<<ans1<<" "<<ans2<<endl;

 }

D

线段树好题

题意：n个女性比三种属性，一旦有一个女的三种属性都被另一个女的压制，那这个女的会自杀，问多少女性会自杀

思路：

先按第一个属性从大到小严格排序，然后再把第二个属性离散化作为线段树的下标，最后再把第三个属性作为线段树上的值，然后查询最大值即可

如果第一个属性有相等的情况，需要把相等的情况全部比较完再更新

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #define lson num<<1,s,mid
 #define rson num<<1|1,mid+1,e
 #define maxn 500005

 using namespace std;

 struct node
 {
     int a,b,c;
     bool operator < (const node &cmp)const
     {
         if(a!=cmp.a)return a<cmp.a;
         if(b!=cmp.b)return b<cmp.b;
         return c<cmp.c;
     }
 }wm[maxn];

 int res[maxn<<];
 int x[maxn];

 void pushup(int num)
 {
     res[num]=max(res[num<<],res[num<<|]);
 }
 void build(int num,int s,int e)
 {
     res[num]=-;
     if(s==e)return ;
     int mid=(s+e)>>;
     build(lson);build(rson);
 }
 void update(int num,int s,int e,int pos,int val)
 {
     if(s==e)
     {
         res[num]=max(res[num],val);
         return;
     }
     int mid=(s+e)>>;
     if(pos<=mid)update(lson,pos,val);
     else update(rson,pos,val);
     pushup(num);
 }
 int query(int num,int s,int e,int l,int r)
 {
     if(l<=s  && r>=e)return res[num];
     int mid=(s+e)>>;
     if(r<=mid)return query(lson,l,r);
     else if(l>mid)return query(rson,l,r);
     else return max(query(lson,l,mid),query(rson,mid+,r));
 }
 int main()
 {
     int n;
     scanf("%d",&n);
     for(int i=;i<=n;i++)
         scanf("%d",&wm[i].a);
     for(int i=;i<=n;i++)
     {
         scanf("%d",&wm[i].b);
         x[i]=wm[i].b;
     }
     for(int i=;i<=n;i++)
         scanf("%d",&wm[i].c);

     sort(wm+,wm++n);

     sort(x+,x++n);

     int m=unique(x+,x++n)-x;
     m--;

     int last=wm[n].a;
     int r=n;
     int l=n;
     int ans=;

     wm[].a=0x3f3f3f3f;

     for(int i=n;i>=;)
     {
         while(wm[l].a==last)
         {
             l--;
         }
         int c=r;
         while(c>l)
         {
             int pos=lower_bound(x+,x+m+,wm[c].b)-x;
             if(pos+<=m && query(,,m,pos+,m)>wm[c].c)ans++;
             c--;
         }
         c=r;
         while(c>l)
         {
             int pos=lower_bound(x+,x+m+,wm[c].b)-x;
             update(,,m,pos,wm[c].c);
             c--;
         }
         i=l;r=l;last=wm[i].a;
     }
     printf("%d\n",ans);
     return ;
 }

E

构造题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int book[][];

 int main(){
     #ifndef ONLINE_JUDGE
       //  freopen("1.txt","r",stdin);
     #endif
     int n;
     std::ios::sync_with_stdio(false);
     cin>>n;
      for(int i  =  ; i < n- ; i++){
          for(int j =  ; j < n- ; j++)
              book[i][j] = (i+j)%(n-)+;
      }
      for(int i =  ; i < n ; i++){
          book[i][n-] = book[i][i];
          book[n-][i] = book[i][i];
          book[i][i] = ;
      }
      for(int i =  ; i < n ; i++){
          for(int j =  ; j < n ; j++)
              cout<<book[i][j]<<" ";
          cout<<endl;
      }

 }