Codeforces Beta Round #46 (Div. 2)
http://codeforces.com/contest/49
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 maxn 1000005 8 typedef long long ll; 9 typedef unsigned long long ull; 10 11 12 int main(){ 13 #ifndef ONLINE_JUDGE 14 freopen("input.txt","r",stdin); 15 #endif 16 std::ios::sync_with_stdio(false); 17 map<char,int>mp; 18 mp['A']++; 19 mp['E']++; 20 mp['I']++; 21 mp['O']++; 22 mp['U']++; 23 mp['Y']++; 24 mp['a']++; 25 mp['e']++; 26 mp['i']++; 27 mp['o']++; 28 mp['u']++; 29 mp['y']++; 30 string str; 31 getline(cin,str); 32 char ch; 33 for(int i=str.length()-1;i>=0;i--){ 34 if((str[i]>='A'&&str[i]<='Z')||(str[i]>='a'&&str[i]<='z')){ 35 ch=str[i]; 36 break; 37 } 38 } 39 if(mp[ch]) cout<<"YES"<<endl; 40 else cout<<"NO"<<endl; 41 }
B
模拟进制转换和运算
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 maxn 1000005 8 typedef long long ll; 9 typedef unsigned long long ull; 10 11 int getmax(int a){ 12 int Max=0; 13 while(a){ 14 Max=max(a%10,Max); 15 a/=10; 16 } 17 return Max; 18 } 19 20 int Change(int n,int base){ 21 int p=1; 22 int ans=0; 23 while(n){ 24 ans=ans+(n%10)*p; 25 n/=10; 26 p*=base; 27 } 28 return ans; 29 } 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 int a,b; 38 cin>>a>>b; 39 int base=max(getmax(a),getmax(b))+1; 40 int sum=Change(a,base)+Change(b,base); 41 int ans=0; 42 while(sum){ 43 sum/=base; 44 ans++; 45 } 46 cout<<ans<<endl; 47 }
C
找规律
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 maxn 1000005 8 typedef long long ll; 9 typedef unsigned long long ull; 10 11 12 13 int main(){ 14 #ifndef ONLINE_JUDGE 15 // freopen("input.txt","r",stdin); 16 #endif 17 std::ios::sync_with_stdio(false); 18 int n; 19 cin>>n; 20 cout<<n<<" "; 21 for(int i=1;i<n;i++) cout<<i<<" "; 22 }
D
枚举第一个是0还是1，然后不断向后遍历判断，取最小值
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 maxn 1000005 8 typedef long long ll; 9 typedef unsigned long long ull; 10 11 int n; 12 string str; 13 14 int func(int ch){ 15 int ans=0; 16 if(str[0]!=ch+'0') ans++; 17 ch^=1; 18 for(int i=1;i<str.length();i++){ 19 if(str[i]!=ch+'0'){ 20 ans++; 21 } 22 ch^=1; 23 } 24 return ans; 25 } 26 27 int main(){ 28 #ifndef ONLINE_JUDGE 29 // freopen("input.txt","r",stdin); 30 #endif 31 std::ios::sync_with_stdio(false); 32 cin>>n; 33 cin>>str; 34 int ans=0x3f3f3f3f; 35 ans=min(ans,func(0)); 36 ans=min(ans,func(1)); 37 cout<<ans<<endl; 38 }
E
区间DP
参考博客：https://blog.csdn.net/zhjchengfeng5/article/details/8201105
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[2]; 14 int n; 15 string str[55]; 16 int dp[55][55]; 17 bool book[2][55][55][55]; 18 int len[2]; 19 20 void Init(int id){ 21 len[id]=s[id].length(); 22 rep(i,0,len[id]){ 23 book[id][i][i][s[id][i]-'a']=1; 24 } 25 rep(L,1,len[id]+1){ 26 int st=0; 27 rep(en,st+L-1,len[id]){ 28 rep(mid,st,en){ 29 rep(i,0,n){ 30 if(book[id][st][mid][str[i][3]-'a']&&book[id][mid+1][en][str[i][4]-'a']){ 31 book[id][st][en][str[i][0]-'a']=1; 32 } 33 } 34 } 35 st++; 36 } 37 } 38 } 39 40 int main(){ 41 #ifndef ONLINE_JUDGE 42 freopen("input.txt","r",stdin); 43 #endif 44 std::ios::sync_with_stdio(false); 45 cin>>s[0]>>s[1]; 46 cin>>n; 47 for(int i=0;i<n;i++){ 48 cin>>str[i]; 49 } 50 Init(0),Init(1); 51 memset(dp,0x3f,sizeof(dp)); 52 dp[0][0]=0; 53 rep(mid0,0,len[0]){ 54 rep(mid1,0,len[1]){ 55 rep(st0,mid0,len[0]){ 56 rep(st1,mid1,len[1]){ 57 rep(ch,0,26){ 58 if(book[0][mid0][st0][ch]&&book[1][mid1][st1][ch]){ 59 dp[st0+1][st1+1]=min(dp[st0+1][st1+1],dp[mid0][mid1]+1); 60 } 61 } 62 } 63 } 64 } 65 } 66 int ans=dp[len[0]][len[1]]; 67 if(ans==0x3f3f3f3f) cout<<-1<<endl; 68 else cout<<ans<<endl; 69 }