1436 twin prime 2, 1436 prime

1436 twin prime 2, 1436 prime
1436 twin prime number 2

 

Time Limit: 2 s space limit: 1000 KB title level: Silver Title Description Description

For example, m = 100, n = 6

It will output all the twin prime numbers with 6 Differences of less than 100: for example,

5 11

7 13

....

83 89

Use a half-width space to distinguish between the output number and the number according to this rule, each row.

Input description Input Description

Enter an integer m in the first row as a range (for example, 100)

In the second row, enter an integer k as the tolerance of the target pair prime number (for example, 6)

Output description Output Description

Each row outputs a pair, and the last row outputs: Total Is :? (? Indicates the total number of such pairs,If not, the output Is Total Is: 0.)

Sample Input Sample Input

Example 1:

50 2

Example 2:

100 90

Example 3:

200 199

Sample output Sample Output

Example 1:

3 5
5 7
11 13
17 19
29 31
41 43
Total Is: 6

Example 2:

7 97
Total Is: 1

Example 3:

Total Is: 0

Data range and prompt Data Size & Hint

M <= 5000

CATEGORY tag Tags click here to expandThis question should have a more optimized algorithm, but I think m <= 5000 ...... let alone brute force!

 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 using namespace std; 5 const int MAXN=10000001; 6 int vis[MAXN]; 7 int bc[MAXN]; 8 int now=1; 9 int main()10 {11     int m,n;12     scanf("%d%d",&m,&n);13     for(int i=2;i<=sqrt(m);i++)14     {15         if(vis[i]==0)16         {17             for(int j=i*i;j<=m;j=j+i)18             {19                 vis[j]=1;20             }21         }22     }23     /*int pre=-1;24     25     for(int i=2;i<=m;i++)26     {27         if(vis[i]==0)28         {29             if(i-pre==n)30             {31                 printf("%d %d\n",pre,i);32                 tot++;33             }34             pre=i;35         }36     }*/37     int tot=0;38     for(int i=2;i<=m;i++)39     {40         if(vis[i]==0)41         {42             bc[now]=i;43             now++;44         }45     }46     for(int i=1;i<=now-1;i++)47     {48         for(int j=1;j<=now-1;j++)49         {50             if(bc[j]-bc[i]==n)51             {52                 printf("%d %d\n",bc[i],bc[j]);53                 tot++;54             }55         }56     }57     printf("Total Is:%d",tot);58     return 0;59 }

 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 using namespace std; 5 const int MAXN=10000001; 6 int vis[MAXN]; 7 int bc[MAXN]; 8 int now=1; 9 int main()10 {11     int m,n;12     scanf("%d%d",&m,&n);13     for(int i=2;i<=sqrt(m);i++)14     {15         if(vis[i]==0)16         {17             for(int j=i*i;j<=m;j=j+i)18             {19                 vis[j]=1;20             }21         }22     }23     /*int pre=-1;24     25     for(int i=2;i<=m;i++)26     {27         if(vis[i]==0)28         {29             if(i-pre==n)30             {31                 printf("%d %d\n",pre,i);32                 tot++;33             }34             pre=i;35         }36     }*/37     int tot=0;38     for(int i=2;i<=m;i++)39     {40         if(vis[i]==0)41         {42             bc[now]=i;43             now++;44         }45     }46     for(int i=1;i<=now-1;i++)47     {48         for(int j=1;j<=now-1;j++)49         {50             if(bc[j]-bc[i]==n)51             {52                 printf("%d %d\n",bc[i],bc[j]);53                 tot++;54             }55         }56     }57     printf("Total Is:%d",tot);58     return 0;59 }