CodeForces-237C- Primes on Interval

E - E
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 237C
Description
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.

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

题意：给出区间a，b，以及数量k，使得x到x+l-1这个区间包含最少k个素数，其中x，l满足不等式a ≤ x ≤ b - l + 1和1 ≤ l ≤ b - a + 1，输出最小的l；
思路：既然和素数有关，一定少不了筛法的思想，同时借用nyoj士兵杀敌一的思想，用数组存下前i位的素数数量，这样只需扫描一次存下来就可以O(1)访问了。接着就是二分查找，大约O(nlogn)的时间复杂度查找到最小的l。

代码

#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<string>
using namespace std;
const int maxn=1000005;
int sum[maxn];//前i位有sum[i]个素数
int is_sushu[maxn];//是素数标记为1，否则标记为0
int a,b,k;//含义如题
void shaifa()
{
    memset(sum,0,sizeof(sum));
    memset(is_sushu,0,sizeof(is_sushu));//不是素数就标记为1
    for(int i=2; i<maxn; i++)
    {
        sum[i]=sum[i-1];
        if(is_sushu[i]==0)//如果是素数
        {
            sum[i]++;//素数数量加一
            for(int j=1; i*j<=maxn; j++) //筛法
                is_sushu[i*j]=1;
        }
    }
}
bool check(int x)//传入二分查找的中间值
{
    for(int i=a; i<=b-x+1; i++)
        if(sum[i+x-1]-sum[i-1]<k)//此区间的素数小于k
            return 0;
    return 1;//满足条件返回1
}
int main()
{
    shaifa();
    while(~scanf("%d%d%d",&a,&b,&k))
    {
        if(sum[b]-sum[a-1]<k)//取极限，如果不满足条件
        {
            printf("-1\n");
            return 0;
        }
        int left=1;//左边界
        int right=b-a+1;//右边界
        int num;
        while(left<=right)//二分查找
        {
            int mid=(left+right)/2;//中间值
            if(check(mid))//满足条件
            {
                num=mid;
                right=mid-1;//缩小范围继续查找
            }
            else
                left=mid+1;//改变左边界
        }
        printf("%d\n",num);
    }
}