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[5];

int main(){
    #ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
    #endif
    for(int i=0;i<3;i++){
        cin>>str[i];
    }
    for(int i=0;i<3;i++){
        for(int j=0;j<3;j++){
            if(str[i][j]!=str[2-i][2-j]){
                cout<<"NO"<<endl;
                return 0;
            }
        }
    }
    cout<<"YES"<<endl;


    return 0;
}

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[5];
int ch[15];

int main(){
    #ifndef ONLINE_JUDGE
     //   freopen("1.txt","r",stdin);
    #endif
    string str1,str2,ans="";
    cin>>str1>>str2;
    for(int i=0;i<str1.length();i++){
        ch[str1[i]-'0']++;
    }
    for(int i=1;i<10;i++){
        if(ch[i]){
            ans+=char(i+'0');
            ch[i]--;
            break;
        }
    }

    for(int i=0;i<10;i++){
        for(int j=0;j<ch[i];j++){
            ans+=char(i+'0');
        }
    }
   // cout<<ans<<" "<<str2<<endl;
    if(ans==str2) cout<<"OK"<<endl;
    else cout<<"WRONG_ANSWER"<<endl;
    return 0;
}

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[105];
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=0;i<n;i++) cin>>a[i];
    for(int i=0;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=0,ans2=0;
    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=0;i<ve.size();i++){
        ans1+=ve[i].first*a[i];
    }
    sort(a,a+n,cmp1);
    for(int i=0;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<<2];
int x[maxn];

void pushup(int num)
{
    res[num]=max(res[num<<1],res[num<<1|1]);
}
void build(int num,int s,int e)
{
    res[num]=-1;
    if(s==e)return ;
    int mid=(s+e)>>1;
    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)>>1;
    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)>>1;
    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+1,r));
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&wm[i].a);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&wm[i].b);
        x[i]=wm[i].b;
    }
    for(int i=1;i<=n;i++)
        scanf("%d",&wm[i].c);

    sort(wm+1,wm+1+n);

    sort(x+1,x+1+n);

    int m=unique(x+1,x+1+n)-x;
    m--;

    int last=wm[n].a;
    int r=n;
    int l=n;
    int ans=0;

    wm[0].a=0x3f3f3f3f;

    for(int i=n;i>=1;)
    {
        while(wm[l].a==last)
        {
            l--;
        }
        int c=r;
        while(c>l)
        {
            int pos=lower_bound(x+1,x+m+1,wm[c].b)-x;
            if(pos+1<=m && query(1,1,m,pos+1,m)>wm[c].c)ans++;
            c--;
        }
        c=r;
        while(c>l)
        {
            int pos=lower_bound(x+1,x+m+1,wm[c].b)-x;
            update(1,1,m,pos,wm[c].c);
            c--;
        }
        i=l;r=l;last=wm[i].a;
    }
    printf("%d
",ans);
    return 0;
}

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[1005][1005];

int main(){
    #ifndef ONLINE_JUDGE
      //  freopen("1.txt","r",stdin);
    #endif
    int n;
    std::ios::sync_with_stdio(false);
    cin>>n;
     for(int i  = 0 ; i < n-1 ; i++){
         for(int j = 0 ; j < n-1 ; j++)
             book[i][j] = (i+j)%(n-1)+1;
     }
     for(int i = 0 ; i < n ; i++){
         book[i][n-1] = book[i][i];
         book[n-1][i] = book[i][i];
         book[i][i] = 0;
     }
     for(int i = 0 ; i < n ; i++){
         for(int j = 0 ; j < n ; j++)
             cout<<book[i][j]<<" ";
         cout<<endl;
     }

}