Codeforces 237C Primes on Interval 【二分】

题目链接：Codeforces 237C Primes on Interval

C. Primes on Interval

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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：让你找到最小的l满足1 <= l <= b - a + 1，使得对于任意的x(a <= x <= b - l + 1)均有[x, x + l - 1]里面至少有k个质数。

思路：推敲下，会发现l是单调的，那么直接二分就好了。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <queue>
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e6 + 1;
const int MOD = 1073741824;
const int INF = 0x3f3f3f3f;
void add(LL &x, LL y) { x += y; x %= MOD; }
bool vis[MAXN];
int sum[MAXN];
void getvis() {
vis[1] = true;
for(int i = 2; i < MAXN; i++) {
if(vis[i]) continue;
for(int j = 2*i; j < MAXN; j += i) {
vis[j] = true;
}
}
sum[0] = 0;
for(int i = 1; i < MAXN; i++) {
sum[i] = sum[i-1] + (vis[i] == false);
}
}
bool judge(int o, int a, int b, int k) {
for(int i = a; i <= b - o + 1; i++) {
if(sum[i+o-1] - sum[i-1] < k) return false;
}
return true;
}
int main()
{
getvis();
int a, b, k;
while(scanf("%d%d%d", &a, &b, &k) != EOF) {
int l = 1, r = b - a + 1;
int ans = -1;
while(r >= l) {
int mid = (l + r) >> 1;
if(judge(mid, a, b, k)) {
ans = mid;
r = mid - 1;
}
else {
l = mid + 1;
}
}
printf("%d\n", ans);
}
return 0;
}