CodeForces 237C Primes on Interval

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_34195441/article/details/53557329

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.

Example
Input
2 4 2
Output
3
Input
6 13 1
Output
4
Input
1 4 3
Output
-1

•题意：给你一个闭区间[a,b]，求一个最小的L，使得在区间[a,b-L+1]内任取一个数x，可以满足在x,x+1,x+2,……,x+L-2,x+L-1内至少包含k个素数。(1<=a,b,k<=10^6)

•考察内容：筛素数、二分
•一边筛素数，一边处理出一个前缀和sum
•sum(i)表示[1,i]中有多少素数
•那么我们每次查询区间[l,r]中有多少素数，直接查sum[r]-sum[l-1]就可以了
•接下去我们按照题意，对答案L进行二分就可以了

#include<stdio.h>
#include<string.h>
int sum[1000010],p[1000010],a,b,k;

void init()
{//筛数
    int i,j;
    for(i=2;i<1000010;i++)
    {
        sum[i]=sum[i-1];
        if(p[i]) continue;
        sum[i]++;
        for(j=i+i;j<1000010;j+=i)
            p[j]=1; 
    }
}
int check(int mid)
{
    for(int i=a;i<=b-mid+1;i++)
    {
        if(sum[i+mid-1]-sum[i-1]<k)
            return 0;
    }
    return 1;
 } 
int main()
{
    init();
    scanf("%d%d%d",&a,&b,&k);
    if(sum[b]-sum[a-1]<k)
    {
        printf("-1\n");
        return 0;
    }
    int l=1,r=b-a+1,ans;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(check(mid))
        {
            ans=mid;r=mid-1;
        }
        else l=mid+1;
    }
    printf("%d\n",ans);
    return 0;
}