『NYIST』第八届河南省ACM竞赛训练赛[正式赛一]－CodeForces 237C，素数打表，二分查找

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

题目意思看了老半天，冷静分析终于摸清了题意，就是求a,b范围内任意一段包含k个素数的最小长度L;用常规方法的话也可以，但是可能会很麻烦；这里可以用一种二分查找，但都是要打表的；

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=1000000+10;
int a,b,k,s[N],ss[N];
void dabiao()
{
    memset(s,0,sizeof(s));
    memset(ss,-1,sizeof(ss));
    ss[0]=ss[1]=0;
    for(int i=2; i<=N; i++)
    {
        s[i]=s[i-1];
        if(ss[i])
            s[i]+=1;//打表；
        if(ss[i])
        {
            if(i>N/i) continue;
            for(int j=i*i; j<=N; j+=i)
                ss[j]=0;//筛法；
        }
    }
}
int cha(int x)
{
    for(int i=a; i<=b-x+1; i++)
        if(s[i+x-1]-s[i-1]<k)
            return 0;
    return 1;
}
int main()
{
    dabiao();
    while(~scanf("%d%d%d",&a,&b,&k))
    {
        if(s[b]-s[a-1]<k)
            printf("-1\n");整个区间内的素数个数；
        else
        {
          int x=0;
            int l=1,r=b-a+1;
            while(l<=r)
            {
                int mid=(l+r)/2;
                if(cha(mid))
                {
                    x=mid;
                    r=mid-1;
                }
                else
                    l=mid+1;
            }
            printf("%d\n",x);
        }
    }
    return 0;
}

其实理清了思路就不会很难；由此可见思路的重要性；