Codeforces Education Round 11

A(模拟＋数学)

题意:在一个数列当中最少添加多少个数可以使它们两两互质，并打印出添加以后的数列

 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>
10 #include <cmath>
11 #include <queue>
12 #include <stack>
13 using namespace std;
14 const int maxn=2020;
15 const int maxm=1000000000;
16 int a[maxn];
17 int gcd(int a,int b)
18 {
19     if(b==0) return a;
20     return gcd(b,a%b);
21 }
22 int n;
23 int main()
24 {
25     while(cin>>n)
26     {
27         for(int i=0;i<n;i++)
28             scanf("%d",&a[i]);
29         vector<int>b;
30         int cnt=0;
31         for(int i=0;i<n-1;i++)
32         {
33             if(gcd(a[i],a[i+1])>=2){
34                 cnt++;
35                 int k;
36                 for(int j=1;j<=maxm;j++){
37                     if(gcd(a[i],j)<2&&gcd(j,a[i+1])<2){
38                         k=j; break;
39                     }
40                 }
41                 b.push_back(a[i]);
42                 b.push_back(k);
43             }
44             else{
45                 b.push_back(a[i]);
46             }
47         }
48         b.push_back(a[n-1]);
49         cout<<cnt<<endl;
50         for(int i=0;i<b.size()-1;i++)
51             cout<<b[i]<<" ";
52         cout<<b[b.size()-1]<<endl;
53     }
54     return 0;
55 }

B(队列模拟）

题意:根据公交车上下车的顺序，打印下车的顺序

分析:用队列直接进行模拟即可

 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>
10 #include <cmath>
11 #include <queue>
12 #include <stack>
13 using namespace std;
14 const int maxn=150;
15 int a[maxn][4];
16 int n,m;
17 int main()
18 {
19     while(cin>>n>>m)
20     {
21         queue<int>que[4];
22         int k=1;
23         int cnt=0;
24         for(;;)
25         {
26             if(cnt>=n) break;
27             if(k>m) break;
28             que[0].push(k);
29             k++;
30             if(k>m) break;
31             que[3].push(k);
32             k++;
33             cnt++;
34         }
35         for(;;)
36         {
37             if(k>m) break;
38             que[1].push(k);
39             k++;
40             if(k>m)  break;
41             que[2].push(k);
42             k++;
43         }
44         vector<int>b;
45         int t;
46         int f1=0,f2=0,f3=0,f4=0;
47         for(;;)
48         {
49             if(que[1].empty()){
50                 f2=1;
51             }else{
52                 t=que[1].front();
53                 que[1].pop();
54                 b.push_back(t);
55             }
56             if(que[0].empty()){
57                 f1=1;
58             }else{
59                 t=que[0].front();
60                 que[0].pop();
61                 b.push_back(t);
62             }
63              if(que[2].empty()){
64                 f3=1;
65             }else{
66                 t=que[2].front();
67                 que[2].pop();
68                 b.push_back(t);
69             }
70              if(que[3].empty()){
71                 f4=1;
72             }else{
73                 t=que[3].front();
74                 que[3].pop();
75                 b.push_back(t);
76             }
77              if(f1&&f2&&f3&&f4) break;
78         }
79         for(int i=0;i<b.size()-1;i++)
80             cout<<b[i]<<" ";
81         cout<<b[b.size()-1]<<endl;
82     }
83     return 0;
84 }