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{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] .
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) .
For each test case, write in one line the numbers of prime k -tuples of distance s .
1 100 200 4 8
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 }
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原文地址：http://www.cnblogs.com/zzuli2sjy/p/5877293.html
