Codeforces Education Round 11

A (analog + math)

Test instructions: Add a minimum number of numbers in a sequence to make them 22 coprime, and print out the added sequence

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <string>5#include <vector>6#include <algorithm>7#include <Set>8#include <map>9#include <bitset>Ten#include <cmath> One#include <queue> A#include <stack> - using namespacestd; - Const intmaxn=2020; the Const intmaxm=1000000000; - intA[MAXN]; - intgcdintAintb) - { +     if(b==0)returnA; -     returnGCD (b,a%b); + } A intN; at intMain () - { -      while(cin>>N) -     { -          for(intI=0; i<n;i++) -scanf"%d",&a[i]); invector<int>b; -         intCnt=0; to          for(intI=0; i<n-1; i++) +         { -             if(GCD (a[i],a[i+1]) >=2){ thecnt++; *                 intK; $                  for(intj=1; j<=maxm;j++){Panax Notoginseng                     if(GCD (A[I],J) <2&AMP;&AMP;GCD (j,a[i+1]) <2){ -K=j; Break; the                     } +                 } A B.push_back (A[i]); the B.push_back (k); +             } -             Else{ $ B.push_back (A[i]); $             } -         } -B.push_back (a[n-1]); thecout<<cnt<<Endl; -          for(intI=0; I<b.size ()-1; i++)Wuyicout<<b[i]<<" "; theCout<<b[b.size ()-1]<<Endl;  -     } Wu     return 0; -}

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B (queue emulation)

Test instructions: According to the order of getting off the bus, print out the order of getting off

Analysis: Direct simulation with the queue can be

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <string>5#include <vector>6#include <algorithm>7#include <Set>8#include <map>9#include <bitset>Ten#include <cmath> One#include <queue> A#include <stack> - using namespacestd; - Const intmaxn= Max; the inta[maxn][4]; - intn,m; - intMain () - { +      while(cin>>n>>m) -     { +queue<int>que[4]; A         intk=1; at         intCnt=0; -          for(;;) -         { -             if(cnt>=n) Break; -             if(k>m) Break; -que[0].push (k); ink++; -             if(k>m) Break; toque[3].push (k); +k++; -cnt++; the         } *          for(;;) $         {Panax Notoginseng             if(k>m) Break; -que[1].push (k); thek++; +             if(k>m) Break; Aque[2].push (k); thek++; +         } -vector<int>b; $         intT; $         intf1=0, f2=0, f3=0, f4=0; -          for(;;) -         { the             if(que[1].empty ()) { -F2=1; Wuyi}Else{ thet=que[1].front (); -que[1].pop (); Wu B.push_back (t); -             } About             if(que[0].empty ()) { $f1=1;  -}Else{ -t=que[0].front (); -que[0].pop (); A B.push_back (t); +             } the              if(que[2].empty ()) { -f3=1;  $}Else{ thet=que[2].front (); theque[2].pop (); the B.push_back (t); the             } -              if(que[3].empty ()) { inf4=1;  the}Else{ thet=que[3].front (); Aboutque[3].pop (); the B.push_back (t); the             } the              if(F1&AMP;&AMP;F2&AMP;&AMP;F3&AMP;&AMP;F4) Break; +         } -          for(intI=0; I<b.size ()-1; i++) thecout<<b[i]<<" ";BayiCout<<b[b.size ()-1]<<Endl; the     } the     return 0; -}

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Codeforces Education Round 11