『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

input

6 13 1

output

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；

巴特西

『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

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