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CodeForces 237C Primes on Interval

2024-07-14 07:18:45   223人阅读

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         else40             l = mid+1;41     }42     printf("%d\n", ans);43 }

 


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