Codeforces Round #661 (Div. 3)

• A. Remove Smallest

• #include<bits/stdc++.h>
  using namespace std;
  #pragma GCC optimize(2)
  typedef long long ll;
  typedef unsigned long long ull;
  typedef long double ld;
  const int N=100;
  int t,n,a[N];
  
  
  int main()
  {
      ios::sync_with_stdio(false);
      cin.tie(0);cout.tie(0);
      cin>>t;
      while(t--) {
          cin>>n;
          for(int i=1;i<=n;++i) cin>>a[i];
          sort(a+1,a+1+n);
          int p=0;
          for(int i=2;i<=n;++i) p=max(p,a[i]-a[i-1]);
          if(p>1) cout<<"NO\n";
          else cout<<"YES\n";
      }
      return 0;
  }
  

• B. Gifts Fixing

• #include<bits/stdc++.h>
  using namespace std;
  #pragma GCC optimize(2)
  typedef long long ll;
  typedef unsigned long long ull;
  typedef long double ld;
  const ll N=100;
  ll n,t,a[N],b[N];
  
  
  int main()
  {
      ios::sync_with_stdio(false);
      cin.tie(0);cout.tie(0);
      cin>>t;
      while(t--){
          cin>>n;
          ll ma=1e9+10,mb=1e9+10;
          for(ll i=1;i<=n;++i){
              cin>>a[i];
              ma=min(ma,a[i]);
          }
          for(ll i=1;i<=n;++i){
              cin>>b[i];
              mb=min(mb,b[i]);
          }
          ll ans=0;
          for(ll i=1;i<=n;++i){
              ll ta=a[i]-ma;
              ll tb=b[i]-mb;
              if(ta>tb){
                  ans+=tb;
                  ans+=abs(ta-tb);
              }else{
                  ans+=ta;
                  ans+=abs(tb-ta);
              }
          }
          cout<<ans<<"\n";
      }
      return 0;
  }
  

• C. Boats Competition

• #include<bits/stdc++.h>
  using namespace std;
  #pragma GCC optimize(2)
  typedef long long ll;
  typedef unsigned long long ull;
  typedef long double ld;
  const int N = 100;
  int t, n, a[N], ans, rec;
  int inq[N];
  
  int main()
  {
      ios::sync_with_stdio(false);
      cin.tie(0); cout.tie(0);
      cin >> t;
      while (t--) {
          cin >> n;
          memset(inq, 0, sizeof(int) * (n + 1));
          for (int i = 1; i <= n; ++i) cin >> a[i];
          sort(a + 1, a + 1 + n);
          //int l=2,r=a[n]*2;
          int s = a[n] + a[n - 1];
          ans = 0;
          for (; s >= 2; s--) {
              int ta = 0;
              memset(inq, 0, sizeof(int) * (n + 1));
              for (int i = 1; i <= n; ++i) {
                  if (inq[i]) continue;
                  int tmp = s - a[i];
                  inq[i] = 1;
                  int j = lower_bound(a + 1, a + 1 + n, tmp) - a ;
                  if (a[i] + a[j] == s) {
                      while (inq[j] && a[i] + a[j] == s) j++;
                      if (!inq[j] && a[i] + a[j] == s) {
                          ta++;
                          inq[j] = 1;
                          inq[i] = 1;
                      }
                  }
              }
              ans = max(ta, ans);
          }
          cout << ans << "\n";
      }
      return 0;
  }
  

• D. Binary String To Subsequences

• #include<bits/stdc++.h>
  using namespace std;
  #pragma GCC optimize(2)
  typedef long long ll;
  typedef unsigned long long ull;
  typedef long double ld;
  int t,n;
  string str;
  const int N=2e5+10;
  int pos[N];
  int rec[N];
  
  int main()
  {
      ios::sync_with_stdio(false);
      cin.tie(0);cout.tie(0);
      cin>>t;
      while(t--){
          cin>>n;
          cin>>str;
          int len=str.size();
          memset(rec,-1,sizeof(int)*(n+1));
          int mcnt,cnt;
          mcnt=cnt=1;
          char w=‘0‘;
          //int flag=0;
          for(int i=0;i<len;++i){
              int j;
              for (j = i; j < len; ++j) {
                  if (str[j] == str[i]) {
                      if(rec[cnt]!=str[j]-w){
                          pos[j]=cnt;
                          rec[cnt]=str[j]-w;
                          cnt++;
                          mcnt=max(mcnt,cnt-1);
                      } else {
                          while(rec[cnt]==str[j]-w) cnt++;
                          pos[j]=cnt;
                          rec[cnt]=str[j]-w;
                          cnt++;
                          mcnt=max(cnt-1,mcnt);
                      }
                  } else {
                      cnt = 1;
                      break;
                  }
              }
              i = j-1;
          }
          cout<<mcnt<<"\n";
          for(int i=0;i<len;++i) cout<<pos[i]<<" ";
          cout<<"\n";
      }
      return 0;
  }
  

前四道都不是很难就不写文字了，看代码就行 EF等空了填坑吧

Codeforces Round #661 (Div. 3) 题解

原文：https://www.cnblogs.com/DrumWashingMachine-Lhy-NoobInCsu/p/13449446.html