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 = 1001000;
#define  inf (1<<29)
// p[i] is i-th prime's position
bool pp[maxn];
int p[maxn] , cnt = 0;
int ss[maxn] , tt[maxn];
void init() {
    cnt = 0;
    pp[0] = pp[1] = 1;
    tt[0] = tt[1] = -1;
    for(int i=2;i<maxn;i++) {
        if(!pp[i]) {
            p[cnt++] = i;
            for(int j=i+i;j<maxn;j+=i) {
                pp[j] = true;
            }
        }
        tt[i] = cnt - 1;
    }
    for(int i=0;i<maxn;i++) {
        if(!pp[i]) ss[i] = tt[i];
        else ss[i] = tt[i] + 1;
    }
}
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 + 1;
        //printf("debug:
");
        //printf("s is %d , t is %d
" , s , t);
        //printf("first pri is %d , last prime is %d
" , p[s] , p[t]);
        if(num < k) {
            printf("-1
");
            continue;
        }
        int ans = 0;
        int tmp;
        tmp = b - p[t-k+1] + 1;
        if(tmp > ans) ans = tmp;
        tmp = p[s+k-1] - a + 1;
        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
" , ans);
    }
    return 0;
}