codeforces - 273C Prime on Interval 【二分】

 

C. Primes on Interval

time limit per test

1 second

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.

Examples

input

2 4 2

output

3

input

6 13 1

output

4

input

1 4 3

output

-1

题意：

 

给你三个数，a，b，k让你求出最小的l满足：对于任意的x属于 (a ≤ x ≤ b - l + 1)中至少含有K个素数。

思路：

二分答案。

#include <bits/stdc++.h>
using namespace std;
bool notprime[1000010];
int num[1000010];
int a, b, k;
void init() {
    notprime[0] = notprime[1] = true;
    for (int i = 2; i <= 1000005; i++) {
        if (notprime[i])   continue;
        for (int j = 2*i;  j < 1000005; j += i) {
            notprime[j] = true;
        }
    }
    num[0] = 0;
    for (int i = 1; i <= 1000005; i++) {
        num[i] = num[i - 1] + !notprime[i];
    }
}
bool judge(int x) {
    for (int i = a; i <= b - x + 1; i++) {
        if (num[i + x - 1] - num[i - 1] < k)
            return false;
    }
    return true;
}
int main() {
    int ans; init();
    while (scanf("%d%d%d", &a, &b, &k) != EOF) {
        ans = -1;
        int lb = 1, ub = b - a + 1;
        while (ub >= lb) {
            int mid = (ub + lb) >> 1;
            if (judge(mid)) {
                ans = mid; ub = mid - 1;
            }
            else lb = mid + 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}