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=;i<s1.length();i++){
             if(s1[i]!=s2[s1.length()-i-]){
                 cout<<"NO"<<endl;
                 return ;
             }
         }
         cout<<"YES"<<endl;
         return ;
     }
     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[];
 int c[][];
 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=;i<=n;i++) cin>>a[i];
     int ans=b;
     priority_queue<int,vector<int>, greater<int> >Q;
     Q.push(a[]);
     for(int i=;i<=n;i++){
         int tmp=b;
         vector<int>ve;
         ve.clear();
         for(int j=;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=;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[];
     int flag=;
     for(int i=;i<str.length();i++){
         if(str[i]=='d'&&i<str.length()-){
             if(str[i+]=='o'&&str[i+]=='t'){
                 cout<<'.';
                 i+=;
             }
             else{
                 cout<<str[i];
             }
         }
         else if(str[i]=='a'){
             if(str[i+]=='t'&&!flag){
                 cout<<'@';
                 flag=;
                 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[];
 bool dp[][][];
 short pre[][][];

 void dfs(int floor,int pos,int v){
     if(floor==n-){
         cout<<pos+<<endl;
         return;
     }
     dfs(floor+,pre[floor][pos][v],v-(str[floor][pos]-''));
     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=;i<n;i++) cin>>str[i];
     for(int i=;i<m;i++){
         dp[n-][i][str[n-][i]-'']=;
     }
     for(int i=n-;i>=;i--){
         for(int j=;j<m;j++){
             for(int k=;k<=;k++){
                 if(dp[i+][j][k]){
                     if(j->=){
                         dp[i][j-][k+str[i][j-]-'']=;
                         pre[i][j-][k+str[i][j-]-'']=j;
                     }
                     if(j+<m){
                         dp[i][j+][k+str[i][j+]-'']=;
                         pre[i][j+][k+str[i][j+]-'']=j;
                     }
                 }
             }
         }
     }
     int Max=-;
     int pos=;
     for(int i=;i<m;i++){
         for(int j=;j>=;j--){
             if((dp[][i][j]==)&&(j%(kk+)==)){
                 if(Max<j){
                     Max=j;
                     pos=i;
                 }
             }
         }
     }
     if(Max==-){
         cout<<-<<endl;
         return ;
     }
     cout<<Max<<endl;
     dfs(,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==){
         cout<<<<endl;
     }
     else if(n==){
         cout<<<<endl;
         cout<<"1 2"<<endl;
     }
     else{
         cout<<(n/)*(n/+n%)<<endl;
         for(int i=;i<=n/;i++){
             for(int j=n/+;j<=n;j++){
                 cout<<i<<" "<<j<<endl;
             }
         }
     }
 }