Codeforces Beta Round #40 (Div. 2)

Codeforces Beta Round #40 (Div. 2)

http://codeforces.com/contest/41

A

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxn 1000006


int main(){
    #ifndef ONLINE_JUDGE
     //   freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    string s1,s2;
    cin>>s1>>s2;
    if(s1.length()==s2.length()){
        for(int i=0;i<s1.length();i++){
            if(s1[i]!=s2[s1.length()-i-1]){
                cout<<"NO"<<endl;
                return 0;
            }
        }
        cout<<"YES"<<endl;
        return 0;
    }
    cout<<"NO"<<endl;
}

 

B

贪心，每种情况枚举过去，买的话建一个优先队列存最小值

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxn 1000006

int a[2005];
int c[2005][2005];
int n,b;

int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    cin>>n>>b;
    for(int i=1;i<=n;i++) cin>>a[i];
    int ans=b;
    priority_queue<int,vector<int>, greater<int> >Q;
    Q.push(a[1]);
    for(int i=2;i<=n;i++){
        int tmp=b;
        vector<int>ve;
        ve.clear();
        for(int j=1;j<i;j++){
            if(!Q.empty()&&tmp>Q.top()){
                ve.push_back(Q.top());
                ans=max(ans,tmp/Q.top()*a[i]+tmp%Q.top());
                tmp%=Q.top();
            }
        }
        Q.push(a[i]);
        for(int j=0;j<ve.size();j++){
            Q.push(ve[j]);
        }
    }
    cout<<ans<<endl;
}

 

C

模拟

#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 pb push_back
#define eps 1e-8
#define PI acos(-1.0)
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull;

/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */


int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    string str;
    cin>>str;
    cout<<str[0];
    int flag=0;
    for(int i=1;i<str.length();i++){
        if(str[i]=='d'&&i<str.length()-3){
            if(str[i+1]=='o'&&str[i+2]=='t'){
                cout<<'.';
                i+=2;
            }
            else{
                cout<<str[i];
            }
        }
        else if(str[i]=='a'){
            if(str[i+1]=='t'&&!flag){
                cout<<'@';
                flag=1;
                i++;
            }
            else{
                cout<<str[i];
            }
        }
        else{
            cout<<str[i];
        }
    }
}

 

D

DP

有句话说的好，DP不行，再加一维。。。

加上的那一维为到该点的值

复杂度为O（n*m*9*n）

注意，0也算是k+1的倍数

#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 pb push_back
#define eps 1e-8
#define PI acos(-1.0)
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull;

/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */
int n,m,kk;
string str[105];
bool dp[105][105][905];
short pre[105][105][905];

void dfs(int floor,int pos,int v){
    if(floor==n-1){
        cout<<pos+1<<endl;
        return;
    }
    dfs(floor+1,pre[floor][pos][v],v-(str[floor][pos]-'0'));
    if(pre[floor][pos][v]<pos){
        cout<<'R';
    }
    else{
        cout<<'L';
    }
}

int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    cin>>n>>m>>kk;
    for(int i=0;i<n;i++) cin>>str[i];
    for(int i=0;i<m;i++){
        dp[n-1][i][str[n-1][i]-'0']=1;
    }
    for(int i=n-2;i>=0;i--){
        for(int j=0;j<m;j++){
            for(int k=0;k<=900;k++){
                if(dp[i+1][j][k]){
                    if(j-1>=0){
                        dp[i][j-1][k+str[i][j-1]-'0']=1;
                        pre[i][j-1][k+str[i][j-1]-'0']=j;
                    }
                    if(j+1<m){
                        dp[i][j+1][k+str[i][j+1]-'0']=1;
                        pre[i][j+1][k+str[i][j+1]-'0']=j;
                    }
                }
            }
        }
    }
    int Max=-1;
    int pos=0;
    for(int i=0;i<m;i++){
        for(int j=900;j>=0;j--){
            if((dp[0][i][j]==1)&&(j%(kk+1)==0)){
                if(Max<j){
                    Max=j;
                    pos=i;
                }
            }
        }
    }
    if(Max==-1){
        cout<<-1<<endl;
        return 0;
    }
    cout<<Max<<endl;
    dfs(0,pos,Max);
}

 

E

找规律，题意说不存在3的循环，所以可以把图看成二分图

边的总数为：(n/2)*(n/2+n%2)

#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 pb push_back
#define eps 1e-8
#define PI acos(-1.0)
#define maxn 1000005
typedef long long ll;
typedef unsigned long long ull;

/*#ifndef ONLINE_JUDGE
        freopen("1.txt","r",stdin);
#endif */

int main(){
    #ifndef ONLINE_JUDGE
      //  freopen("input.txt","r",stdin);
    #endif
    std::ios::sync_with_stdio(false);
    int n;
    cin>>n;
    if(n==1){
        cout<<0<<endl;
    }
    else if(n==2){
        cout<<1<<endl;
        cout<<"1 2"<<endl;
    }
    else{
        cout<<(n/2)*(n/2+n%2)<<endl;
        for(int i=1;i<=n/2;i++){
            for(int j=n/2+1;j<=n;j++){
                cout<<i<<" "<<j<<endl;
            }
        }
    }
}