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
2 4 2
3
6 13 1
4
1 4 3
-1
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 }
CodeForces 237C Primes on Interval,布布扣,bubuko.com
CodeForces 237C Primes on Interval
原文：http://www.cnblogs.com/zzy9669/p/3871588.html