codeforces 237C. Primes 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'
4000
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：给出三个正整数a、b、k，求最小的L（1 ≤ L≤ b - a + 1）满足对于[a, b-L+1]中的任意一个数X，在[X, X+L-1]这L个数中，至少有k个素数。如果不存在满足条件的L，输出-1.

解题思路：首先判断是否有解，即判断当L=b-a+1时是否有解，因此时L最大，若此时都无解，则肯定无解。若有解，则1 ≤ L≤ b - a + 1，二分求最小的L即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e6 + 10;
int preim[maxn],cnt[maxn];
bool vis[maxn];
void get_prime()
{
memset(vis,false,sizeof(vis));
vis[0] = true;
vis[1] = true;
for(int i = 2; i <= maxn; i++)
{
if(vis[i] == true)
continue;
for(int j = 2 * i; j <= maxn; j += i)
vis[j] = true;
}
cnt[0] = 0;
for(int i = 1; i <= 1000000; i++)
{
if(vis[i] == false)
cnt[i] = cnt[i-1] + 1;
else
cnt[i] = cnt[i-1];
}
}
bool judge(int x,int y,int k,int l)
{
int r = y - l + 1;
for(int i = x; i <= r; ++i)
{
if(cnt[i+l-1] - cnt[i-1] >= k)
continue;
else
return false;
}
return true;
}
int get_ans(int a,int b,int k)
{
int L = 1, R = b - a + 1, mid;
while(L <= R)
{
mid = (L + R) / 2;
if(judge(a,b,k,mid))
R = mid - 1;
else
L = mid + 1;
}
return L;
}
int main()
{
get_prime();
int a,b,k;
while(~scanf("%d%d%d",&a,&b,&k))
{
if(!judge(a,b,k,b - a + 1))
printf("-1\n");
else
printf("%d\n",get_ans(a,b,k));
}
return 0;
}