Educational Codeforces Round 82 (Rated for Div. 2) A-E代码（暂无记录题解）

A. Erasing Zeroes （模拟）

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5+;
 int main(){
     int t;cin>>t;
     while(t--){
         string s;cin>>s;
         bool f1 = ,f2 = ;
         int cnt = ;
         int ans = ;
         for(int i = ;i<s.length();i++){
             if(s[i] == ''){
                 if(f1 == ) {
                     ans+=cnt;
                     cnt = ;
                     f1 = ;
                 }
                 if(f1 == ) f1 = ;
             }
             else{
                 if(f1 == ) cnt++;
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }

B. National Project （数学题 周期计算）

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5+;
 int main(){
     int t;cin>>t;
     while(t--){
         ll n,g,b;
         cin>>n>>g>>b;
         ll tn = n;
         if(n% == ) n = n/+;
         else n = n/;
         if(n<=g ) {
             cout<<tn<<endl;continue;
         }
         else{
             ll t = g+b;//一个周期
             ll c = n/g;//至少需要几个周期
             if(n%g == ) {
                 if(c*t-b<tn) cout<<tn<<endl;
                 else cout<<c*t-b<<endl;
                 continue;
             }
             ll d = n%g;
             ll ans = c*t+d;
             if(ans<tn) ans+=(tn-ans);
             cout<<ans<<endl;
             continue;
         }
         // 4 1 1
     }
     return ;
 }

C. Perfect Keyboard （dfs 图论）

 #include<bits/stdc++.h>
 using namespace std;
 struct node{
     vector<int> v;
 }g[];
 string ans;
 int vis[];
 void init(){
     for(int i = ;i<;i++) g[i].v.clear(),vis[i] = ;
     ans = "";
 }
 void dfs(int cur){
     vis[cur] = ,ans+='a'+cur;
     for(int i = ;i<g[cur].v.size();i++){
         if(!vis[g[cur].v[i]]) dfs(g[cur].v[i]);
     }
 }
 void solve(){
     string s;
     cin>>s;
     init();
     map<pair<int,int>,int > m;
     for(int i = ;i<s.length();i++){
         int a = s[i-] - 'a',b = s[i] - 'a';
         if(a<b) swap(a,b);
         if(m[{a,b}] == ){
              m[{a,b}] = ;
              g[a].v.push_back(b);
              g[b].v.push_back(a);
         }
     }
     for(int i = ;i<;i++){
         if(g[i].v.size()>) {
              cout<<"NO"<<endl;
              return;
         }
     }
     for(int i = ;i<;i++){
         if(!vis[i] && g[i].v.size()<){
             dfs(i);
         }
     }
     if(ans.length() == ) {
         cout<<"YES"<<endl;
         cout<<ans<<endl;
         return;
     }
     else{
         cout<<"NO"<<endl;
     }
 }
 int main(){
     int t;
     cin>>t;
     while(t--){
         solve();
     }
     return ;
 }

D. Fill The Bag（贪心 二进制位运算 状态压缩）

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5+;
 const int maxBit = ;
 int cnt[maxBit];
 void solve(){
     ll n;int m;
     scanf("%lld%d",&n,&m);
     memset(cnt,,sizeof(cnt));
     ll sum = ;
     for(int i = ;i<=m;i++) {
         ll x;
         scanf("%lld",&x);
         sum+=x;
         for(int j = ;j<=maxBit;j++){
             if((x>>j)& == ) {
                 cnt[j+]++;
                 break;
             }
         }
     }
     if(sum<n) {
         cout<<-<<endl;return ;
     }
     int ans = ;
     for(int i = ;i<=maxBit;i++){
         if((n>>(i-))&){
             if(!cnt[i]){
                 int indx = -;
                 for(int j = i;j<=maxBit;j++){
                     if(cnt[j]) {
                         indx = j;
                         break;
                     }
                 }
                 while(indx!=i){
                     cnt[indx]--,cnt[indx-]+=,ans++,indx--;
                 }
 //                n^=(1<<(i-1));
             }
             cnt[i]--;
         }
         cnt[i+] += cnt[i]/;
     }
     cout<<ans<<endl;
 }
 //
 int main(){
     int t;
     cin>>t;
     while(t--){
         solve();
     }
     return ;
 } 

E. Erase Subsequences （字符串上dp）

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 int dp[maxn][maxn];
 bool check(string s,string t){
     int indx = ;
     for(int i = ;i<s.length();i++){
         if(t[indx] == s[i]) indx++;
     }
     if(indx == t.length()) return true;
     return false;
 }
 bool check(string s,string t1,string t2){
     memset(dp,-,sizeof(dp));
     dp[][] = ;
     for(int i = ;i<s.length();i++){
         for(int j = ;j<=t1.size();j++){
             if(dp[i][j]<) continue;
             if(j<t1.size() && s[i] == t1[j]){
                 dp[i+][j+] = max(dp[i+][j+],dp[i][j]);
             }
             if(dp[i][j]<t2.size() && s[i] == t2[dp[i][j]]){
                 dp[i+][j] = max(dp[i+][j],dp[i][j]+);
             }
             dp[i+][j] = max(dp[i+][j],dp[i][j]);
         }
     }
     if(dp[s.length()][t1.length()] == t2.length()) return true;
     return false;
 }
 void solve(){
     string s,t;
     cin>>s>>t;
     if(check(s,t)){
         cout<<"YES"<<endl;
         return;
     }
     for(int i = ;i<t.length()-;i++){
         string t1 = t.substr(,i+);
         string t2 = t.substr(i+,t.size());
         if(check(s,t1,t2)){
             cout<<"YES"<<endl;
             return;
         }
     }
     cout<<"NO"<<endl;
     return;
 }
 int main(){
     int t;
     cin>>t;
     while(t--){
         solve();
     }
     return ;
 }