CodeForces 237C Primes on Interval

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

•题意：给你一个闭区间[a,b]，求一个最小的L，使得在区间[a,b-L+1]内任取一个数x，可以满足在x,x+1,x+2,……,x+L-2,x+L-1内至少包含k个素数。(1<=a,b,k<=10^6)

 

•考察内容：筛素数、二分

•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 <stdio.h>

 2 #include <string.h>

 3 #define maxn 1000010

 4 

 5 int sum[maxn], a, b, k;

 6 bool pri[maxn];

 7 

 8 void init(){//筛法 素数打表

 9     for(int i = 2; i < maxn; i++){

10         sum[i] = sum[i-1];

11         if(pri[i])

12             continue;

13         sum[i]++;

14         for(int j = i+i; j < maxn; j += i)

15             pri[j] = 1;

16     }

17 }

18 

19 bool check(int mid){

20     for(int i = a; i <= b-mid+1; i++){

21         if(sum[i+mid-1] - sum[i-1] < k)

22             return 0;

23     }

24     return 1;

25 }

26 

27 int main(){

28     init();

29     scanf("%d%d%d", &a, &b, &k);

30     if(sum[b] - sum[a-1] < k){

31         printf("-1\n");

32         return 0;

33     }

34     int l = 1, r = b-a+1, ans;

35     while(l <= r){

36         int mid = (l+r)>>1;

37         if(check(mid))

38             ans = mid, r = mid-1;

39         else

40             l = mid+1;

41     }

42     printf("%d\n", ans);

43 }