codeforce --- 237C

C. Primes on Interval

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 integer x(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 test(s)

input

2 4 2

output

3

input

6 13 1

output

4

input

1 4 3

output

-1

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 const int maxn = ;
 #define  inf (1<<29)
 // p[i] is i-th prime's position
 bool pp[maxn];
 int p[maxn] , cnt = ;
 int ss[maxn] , tt[maxn];
 void init() {
     cnt = ;
     pp[] = pp[] = ;
     tt[] = tt[] = -;
     for(int i=;i<maxn;i++) {
         if(!pp[i]) {
             p[cnt++] = i;
             for(int j=i+i;j<maxn;j+=i) {
                 pp[j] = true;
             }
         }
         tt[i] = cnt - ;
     }
     for(int i=;i<maxn;i++) {
         if(!pp[i]) ss[i] = tt[i];
         else ss[i] = tt[i] + ;
     }
 }
 int main() {
     init();
     int a , b , k;
     while(~scanf("%d%d%d" , &a,&b,&k)) {
         int s = ss[a] , t = tt[b];
         int num = t - s + ;
         //printf("debug:\n");
         //printf("s is %d , t is %d\n" , s , t);
         //printf("first pri is %d , last prime is %d\n" , p[s] , p[t]);
         if(num < k) {
             printf("-1\n");
             continue;
         }
         int ans = ;
         int tmp;
         tmp = b - p[t-k+] + ;
         if(tmp > ans) ans = tmp;
         tmp = p[s+k-] - a + ;
         if(tmp > ans) ans = tmp;
         for(int i=s;i+k<=t;i++) {
             tmp = p[i+k] - p[i];
             if(tmp > ans) ans = tmp;
         }
         printf("%d\n" , ans);
     }
     return ;
 }