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+5;
int main(){
    int t;cin>>t;
    while(t--){
        string s;cin>>s;
        bool f1 = 0,f2 = 0;
        int cnt = 0;
        int ans = 0;
        for(int i = 0;i<s.length();i++){
            if(s[i] == '1'){
                if(f1 == 1) {
                    ans+=cnt;
                    cnt = 0;
                    f1 = 1;
                }
                if(f1 == 0) f1 = 1;
            }
            else{
                if(f1 == 1) cnt++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

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+5;
int main(){
    int t;cin>>t;
    while(t--){
        ll n,g,b;
        cin>>n>>g>>b;
        ll tn = n;
        if(n%2 == 1) n = n/2+1;
        else n = n/2;
        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 == 0) {
                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 0;
}

C. Perfect Keyboard （dfs 图论）

#include<bits/stdc++.h>
using namespace std;
struct node{
    vector<int> v;
}g[26];
string ans;
int vis[26];
void init(){
    for(int i = 0;i<26;i++) g[i].v.clear(),vis[i] = 0;
    ans = ""; 
}
void dfs(int cur){
    vis[cur] = 1,ans+='a'+cur;
    for(int i = 0;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 = 1;i<s.length();i++){
        int a = s[i-1] - 'a',b = s[i] - 'a';
        if(a<b) swap(a,b);
        if(m[{a,b}] == 0){
             m[{a,b}] = 1;
             g[a].v.push_back(b);
             g[b].v.push_back(a);
        }
    }
    for(int i = 0;i<26;i++){
        if(g[i].v.size()>2) {
             cout<<"NO"<<endl;
             return;
        }
    }
    for(int i = 0;i<26;i++){
        if(!vis[i] && g[i].v.size()<2){
            dfs(i);
        }
    }
    if(ans.length() == 26) {
        cout<<"YES"<<endl;
        cout<<ans<<endl;
        return;
    }
    else{
        cout<<"NO"<<endl;
    }
}
int main(){
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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+5;
const int maxBit = 62;
int cnt[maxBit];
void solve(){
    ll n;int m;
    scanf("%lld%d",&n,&m);
    memset(cnt,0,sizeof(cnt));
    ll sum = 0;
    for(int i = 1;i<=m;i++) {
        ll x;
        scanf("%lld",&x);
        sum+=x;
        for(int j = 0;j<=maxBit;j++){
            if((x>>j)&1 == 1) {
                cnt[j+1]++;
                break;
            }
        } 
    }
    if(sum<n) {
        cout<<-1<<endl;return ;
    }
    int ans = 0;
    for(int i = 1;i<=maxBit;i++){
        if((n>>(i-1))&1){
            if(!cnt[i]){
                int indx = -1;
                for(int j = i;j<=maxBit;j++){
                    if(cnt[j]) {
                        indx = j;
                        break;
                    }
                }
                while(indx!=i){
                    cnt[indx]--,cnt[indx-1]+=2,ans++,indx--;
                }
//                n^=(1<<(i-1));            
            }
            cnt[i]--;
        }
        cnt[i+1] += cnt[i]/2;
    }
    cout<<ans<<endl;
}
//
int main(){
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
} 

E. Erase Subsequences （字符串上dp）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 405;
int dp[maxn][maxn];
bool check(string s,string t){
    int indx = 0;
    for(int i = 0;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,-1,sizeof(dp));
    dp[0][0] = 0;
    for(int i = 0;i<s.length();i++){
        for(int j = 0;j<=t1.size();j++){
            if(dp[i][j]<0) continue;
            if(j<t1.size() && s[i] == t1[j]){
                dp[i+1][j+1] = max(dp[i+1][j+1],dp[i][j]);
            }
            if(dp[i][j]<t2.size() && s[i] == t2[dp[i][j]]){
                dp[i+1][j] = max(dp[i+1][j],dp[i][j]+1);
            }
            dp[i+1][j] = max(dp[i+1][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 = 0;i<t.length()-1;i++){
        string t1 = t.substr(0,i+1);
        string t2 = t.substr(i+1,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 0;
}