{p1,..., pk : p1 < p2 <...< pk} is called a prime k -tuple of distance s if p1, p2,..., pk are consecutive prime numbers and pk - 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 }