3998 - Prime k-tuple

{p₁,..., p_k : p₁ < p₂ <...< p_k} is called a prime k -tuple of distance s if p₁, p₂,..., p_k are consecutive prime numbers and p_k - p₁ = s . For example, with k = 4 , s = 8 , {11, 13, 17, 19} is a prime 4-tuple of distance 8.

Given an interval [a, b] , k , and s , your task is to write a program to find the number of prime k -tuples of distance s in the interval [a, b] .

Input

The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.

For each data set, there is only one line containing 4 numbers, a , b , k and s (a, b < 2 * 10⁹, k < 10, s < 40) .

Output

For each test case, write in one line the numbers of prime k -tuples of distance s .

Sample Input

1
100 200 4 8

Sample Output

2

思路：区间筛素数；

数据就是很水，没给定【a,b】区间的大小，然后筛法水过。然后统计的时候，维护两个端点就行了。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<queue>
 6 #include<math.h>
 7 using namespace std;
 8 typedef long long LL;
 9 bool prime[100005];
10 bool prime_max[20*1000000];
11 int ans[100005];
12 LL ac[1000007];
13 int  main(void)
14 {
15         memset(prime,0,sizeof(prime));
16         int i,j;
17         int cn = 0;
18         for(i = 2; i < 1000; i++)
19                 if(!prime[i])
20                         for(j = i; (i*j) < 100005; i++)
21                                 prime[i*j] = true;
22         for(i = 2; i < 100005; i++)
23         {
24                 if(!prime[i])
25                         ans[cn++] = i;
26         }
27         int n;
28         scanf("%d",&n);
29         LL a,b,k,t;
30         while(n--)
31         {
32                 scanf("%lld %lld %lld %lld",&a,&b,&k,&t);
33                 memset(prime_max,0,sizeof(prime_max));
34                 LL s ;
35                 for(i = 0; i < cn; i++)
36                 {
37                         for(s = max(2LL,a/ans[i])*ans[i]; s <= b; s += ans[i])
38                         {
39                                 if(s >= a)
40                                         prime_max[s-a] = true;
41                         }
42                 }
43                 int v = 0;
44                 for(s = a; s <= b; s++)
45                 {
46                         if(!prime_max[s-a])
47                                 ac[v++] = s;
48                 }
49                 int l = 0;
50                 int r = k-1;
51                 LL ask = 0;
52                 while(true)
53                 {
54                         if(r >= v)
55                                 break;
56                         if(ac[r] - ac[l] == t)
57                         {
58                                 ask++;
59                         }
60                         l++;
61                         r++;
62                 }
63                 printf("%lld\n",ask);
64         }
65         return 0;
66 }