Codeforces Beta Round #57 (Div. 2)

Codeforces Beta Round #57 (Div. 2)

http://codeforces.com/contest/61

A

 #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 eb emplace_back
 #define maxn 1000005
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 string s1,s2;

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     cin>>s1>>s2;
     rep(i,,s1.length()){
         if(s1[i]==s2[i]){
             cout<<;
         }
         else cout<<;
     }
 }

B

 #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 eb emplace_back
 #define maxn 1000005
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 string s[];
 map<string,int>mp;

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     string str;
     rep(i,,){
          s[i]="";
          cin>>str;
          rep(j,,str.length()){
             if(str[j]>='A'&&str[j]<='Z') str[j]+=;
             if(str[j]>='a'&&str[j]<='z') s[i]+=str[j];
          }
     }
     int n;
     cin>>n;
     string ss;
     mp[s[]]=,mp[s[]]=,mp[s[]]=;
     mp[s[]+s[]]=;mp[s[]+s[]]=;mp[s[]+s[]]=;mp[s[]+s[]]=;mp[s[]+s[]]=;mp[s[]+s[]]=;
     mp[s[]+s[]+s[]]=;mp[s[]+s[]+s[]]=;mp[s[]+s[]+s[]]=;mp[s[]+s[]+s[]]=;mp[s[]+s[]+s[]]=;mp[s[]+s[]+s[]]=;
     while(n--){
         cin>>str;
         ss="";
         rep(i,,str.length()){
             if(str[i]>='A'&&str[i]<='Z') str[i]+=;
             if(str[i]>='a'&&str[i]<='z') ss+=str[i];
         }
         if(mp[ss]) cout<<"ACC"<<endl;
         else cout<<"WA"<<endl;
     }
 }

C

进制模拟

 #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 eb emplace_back
 #define maxn 1000005
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 string a[]={"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
 int v[]={,,,,,,,,,,,,};
 string val[];

 string roman(int n) {
     string ret="";
     for(int i=;i<;i++)
         while(n>=v[i]) {
             n-=v[i];
             ret += a[i];
         }
     return ret;
 }
 void pre() {
     for(int i=;i<=;i++)
         val[i] = roman(i);
 }

 char c[], B[];
 int A;

 int main(){
     #ifndef ONLINE_JUDGE
        // freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     pre();
     cin>>A>>B>>c;
     long long v = ;
     for(int i=;c[i];i++) {
         v=v*A + (c[i]>='A'? c[i]-'A'+: c[i]-'');
     }
     if(B[]=='R') cout<<val[v].c_str()<<endl;
     else {
         char d[];
         int dn=, base = atoi(B);
         if(v==) {
             d[dn++]=;
         }
         while(v>) {
             d[dn++]=v%base;
             v/=base;
         }
         while(dn-->) cout<<char(d[dn]>=? d[dn]-+'A': d[dn]+'');
         cout<<endl;
     }
 }

D

通过模拟找规律可以发现，最短距离为所有的距离之和*2减去根结点下的最长链

 #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 eb emplace_back
 #define maxn 1000005
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 int n;
 vector<pair<int,ll> >ve[];

 int dfs(int pos,int pre){
     ll Max=;
     ll tmp;
     ll now;
     rep(i,,ve[pos].size()){
         if(ve[pos][i].first!=pre){
             now=ve[pos][i].second;
             tmp=dfs(ve[pos][i].first,pos);
             if(tmp+now>Max){
                 Max=tmp+now;
             }
         }
     }
     return Max;
 }

 int main(){
     #ifndef ONLINE_JUDGE
        // freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     cin>>n;
     int u,v;
     ll w;
     ll sum=;
     rep(i,,n){
         cin>>u>>v>>w;
         sum+=w;
         ve[u].pb(make_pair(v,w));
         ve[v].pb(make_pair(u,w));
     }
     sum+=sum;
     int Max=;
     ll tmp;
     rep(i,,ve[].size()){
         tmp=dfs(ve[][i].first,);
         if(tmp+ve[][i].second>Max){
             Max=tmp+ve[][i].second;
         }
     }
     cout<<sum-Max<<endl;
 }

E

找三元祖，问前面比a[i]大且后面比a[i]小的三元组有多少个

直接上树状树组，离散化后枚举每个点，找出前面比它大的数，然后乘上后面比它小的数即可，会爆int

 #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 eb emplace_back
 #define maxn 1000006
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 int tree[maxn];

 int lowbit(int x){
     return x&(-x);
 }

 void update(int x){
     while(x<maxn){
         tree[x]+=;
         x+=lowbit(x);
     }
 }

 int query(int x){
     int ans=;
     while(x){
         ans+=tree[x];
         x-=lowbit(x);
     }
     return ans;
 }

 int a[maxn];
 int n;
 vector<int>ve;

 int getpos(int x){
     return lower_bound(ve.begin(),ve.end(),x)-ve.begin()+;
 }

 ll pre[maxn];

 int main(){
     #ifndef ONLINE_JUDGE
        // freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     cin>>n;
     rep(i,,n+){
         cin>>a[i];
         ve.pb(a[i]);
     }
     sort(ve.begin(),ve.end());
     ve.erase(unique(ve.begin(),ve.end()),ve.end());
     int pos;
     update(getpos(a[]));
     ll ans=;
     rep(i,,n){
         pos=getpos(a[i]);
         pre[i]=query(maxn-)-query(pos);
      //   cout<<query(maxn-1)<<" "<<query(pos)<<endl;
         update(pos);
         ///
       //  cout<<pre[i]<<" "<<((n-i)-(n-pos-pre[i]))<<endl;
         ans+=pre[i]*(-i+pos+pre[i]);
     }
     cout<<ans<<endl;
 }