Description
You‘ve decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.
Consider positive integers a, a + 1, ..., b (a ≤ b). You want to find the minimum integer l (1 ≤ l ≤ b - a + 1) such that for any integerx (a ≤ x ≤ b - l + 1) among l integers x, x + 1, ..., x + l - 1 there are at least k prime numbers.
Find and print the required minimum l. If no value l meets the described limitations, print -1.
Input
A single line contains three space-separated integers a, b, k (1 ≤ a, b, k ≤ 106; a ≤ b).
Output
In a single line print a single integer — the required minimum l. If there‘s no solution, print -1.
Sample Input
Input
2 4 2
Output
3
Input
6 13 1
Output
4
Input
1 4 3
Output
-1
一边筛素数，一边处理出一个前缀和sum sum(i)表示[1,i]中有多少素数 那么我们每次查询区间[l,r]中有多少素数，直接查sum[r]-sum[l-1]就可以了 接下去我们按照题意，对答案L进行二分就可以了
1 #include <iostream>
2 using namespace std;
3
4 const int maxn = 1000001;
5 int sum[maxn],a,b,k;
6 bool pri[maxn];
7 void init(){
8 for(int i = 2;i < maxn;i++){
9 sum[i] = sum[i-1];
10 if(pri[i]) continue;
11 sum[i]++;
12 for(int j = i+i;j < maxn;j += i)
13 pri[j] = 1;
14 }
15 }
16
17 bool check(int mid){
18 for(int i = a;i <= b-mid+1;i++){
19 if(sum[i+mid-1]-sum[i-1] < k) return 0;
20 }
21 return 1;
22 }
23
24 int main(){
25 init();
26 cin.sync_with_stdio(false);
27 cin>>a>>b>>k;
28 if(sum[b]-sum[a-1] < k){
29 cout<<"-1"<<endl;
30 return 0;
31 }
32 int l = 1,r = b-a+1,ans;
33 while(l <= r){
34 int mid = (l+r)>>1;
35 if(check(mid)) ans = mid,r = mid-1;
36 else l = mid+1;
37 }
38 cout<<ans<<endl;
39 }
Problem E CodeForces 237C