Codeforces Beta Round #57 (Div. 2)
http://codeforces.com/contest/61
A
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 string s1,s2; 14 15 int main(){ 16 #ifndef ONLINE_JUDGE 17 freopen("input.txt","r",stdin); 18 #endif 19 std::ios::sync_with_stdio(false); 20 cin>>s1>>s2; 21 rep(i,0,s1.length()){ 22 if(s1[i]==s2[i]){ 23 cout<<0; 24 } 25 else cout<<1; 26 } 27 }
B
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 string s[3]; 14 map<string,int>mp; 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 freopen("input.txt","r",stdin); 19 #endif 20 std::ios::sync_with_stdio(false); 21 string str; 22 rep(i,0,3){ 23 s[i]=""; 24 cin>>str; 25 rep(j,0,str.length()){ 26 if(str[j]>='A'&&str[j]<='Z') str[j]+=32; 27 if(str[j]>='a'&&str[j]<='z') s[i]+=str[j]; 28 } 29 } 30 int n; 31 cin>>n; 32 string ss; 33 mp[s[0]]=1,mp[s[1]]=1,mp[s[2]]=1; 34 mp[s[0]+s[1]]=1;mp[s[0]+s[2]]=1;mp[s[1]+s[0]]=1;mp[s[1]+s[2]]=1;mp[s[2]+s[0]]=1;mp[s[2]+s[1]]=1; 35 mp[s[0]+s[1]+s[2]]=1;mp[s[0]+s[2]+s[1]]=1;mp[s[1]+s[0]+s[2]]=1;mp[s[1]+s[2]+s[0]]=1;mp[s[2]+s[1]+s[0]]=1;mp[s[2]+s[0]+s[1]]=1; 36 while(n--){ 37 cin>>str; 38 ss=""; 39 rep(i,0,str.length()){ 40 if(str[i]>='A'&&str[i]<='Z') str[i]+=32; 41 if(str[i]>='a'&&str[i]<='z') ss+=str[i]; 42 } 43 if(mp[ss]) cout<<"ACC"<<endl; 44 else cout<<"WA"<<endl; 45 } 46 }
C
进制模拟
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 string a[13]={"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}; 14 int v[13]={1000,900,500,400,100,90,50,40,10,9,5,4,1}; 15 string val[3005]; 16 17 string roman(int n) { 18 string ret=""; 19 for(int i=0;i<13;i++) 20 while(n>=v[i]) { 21 n-=v[i]; 22 ret += a[i]; 23 } 24 return ret; 25 } 26 void pre() { 27 for(int i=1;i<=3000;i++) 28 val[i] = roman(i); 29 } 30 31 char c[2000], B[10]; 32 int A; 33 34 int main(){ 35 #ifndef ONLINE_JUDGE 36 // freopen("input.txt","r",stdin); 37 #endif 38 std::ios::sync_with_stdio(false); 39 pre(); 40 cin>>A>>B>>c; 41 long long v = 0; 42 for(int i=0;c[i];i++) { 43 v=v*A + (c[i]>='A'? c[i]-'A'+10: c[i]-'0'); 44 } 45 if(B[0]=='R') cout<<val[v].c_str()<<endl; 46 else { 47 char d[2000]; 48 int dn=0, base = atoi(B); 49 if(v==0) { 50 d[dn++]=0; 51 } 52 while(v>0) { 53 d[dn++]=v%base; 54 v/=base; 55 } 56 while(dn-->0) cout<<char(d[dn]>=10? d[dn]-10+'A': d[dn]+'0'); 57 cout<<endl; 58 } 59 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000005 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int n; 14 vector<pair<int,ll> >ve[100005]; 15 16 int dfs(int pos,int pre){ 17 ll Max=0; 18 ll tmp; 19 ll now; 20 rep(i,0,ve[pos].size()){ 21 if(ve[pos][i].first!=pre){ 22 now=ve[pos][i].second; 23 tmp=dfs(ve[pos][i].first,pos); 24 if(tmp+now>Max){ 25 Max=tmp+now; 26 } 27 } 28 } 29 return Max; 30 } 31 32 int main(){ 33 #ifndef ONLINE_JUDGE 34 // freopen("input.txt","r",stdin); 35 #endif 36 std::ios::sync_with_stdio(false); 37 cin>>n; 38 int u,v; 39 ll w; 40 ll sum=0; 41 rep(i,1,n){ 42 cin>>u>>v>>w; 43 sum+=w; 44 ve[u].pb(make_pair(v,w)); 45 ve[v].pb(make_pair(u,w)); 46 } 47 sum+=sum; 48 int Max=0; 49 ll tmp; 50 rep(i,0,ve[1].size()){ 51 tmp=dfs(ve[1][i].first,1); 52 if(tmp+ve[1][i].second>Max){ 53 Max=tmp+ve[1][i].second; 54 } 55 } 56 cout<<sum-Max<<endl; 57 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int tree[maxn]; 14 15 int lowbit(int x){ 16 return x&(-x); 17 } 18 19 void update(int x){ 20 while(x<maxn){ 21 tree[x]+=1; 22 x+=lowbit(x); 23 } 24 } 25 26 int query(int x){ 27 int ans=0; 28 while(x){ 29 ans+=tree[x]; 30 x-=lowbit(x); 31 } 32 return ans; 33 } 34 35 int a[maxn]; 36 int n; 37 vector<int>ve; 38 39 int getpos(int x){ 40 return lower_bound(ve.begin(),ve.end(),x)-ve.begin()+1; 41 } 42 43 ll pre[maxn]; 44 45 int main(){ 46 #ifndef ONLINE_JUDGE 47 // freopen("input.txt","r",stdin); 48 #endif 49 std::ios::sync_with_stdio(false); 50 cin>>n; 51 rep(i,1,n+1){ 52 cin>>a[i]; 53 ve.pb(a[i]); 54 } 55 sort(ve.begin(),ve.end()); 56 ve.erase(unique(ve.begin(),ve.end()),ve.end()); 57 int pos; 58 update(getpos(a[1])); 59 ll ans=0; 60 rep(i,2,n){ 61 pos=getpos(a[i]); 62 pre[i]=query(maxn-1)-query(pos); 63 // cout<<query(maxn-1)<<" "<<query(pos)<<endl; 64 update(pos); 65 /// 66 // cout<<pre[i]<<" "<<((n-i)-(n-pos-pre[i]))<<endl; 67 ans+=pre[i]*(-i+pos+pre[i]); 68 } 69 cout<<ans<<endl; 70 }