codeforces 237 C. Primes on Interval

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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

题解：素数筛+前缀和求值sum[i]+二分。

代码：

#include <iostream>#include <cstdio>#include <cstring>#define maxn 1000010using namespace std;int su[maxn],sum[maxn];int a,b,k;void Prime(){    memset(su,0,sizeof(su));    su[0]=1;    su[1]=1;    for(int i=2;i<maxn;i++)    {        if(su[i])            continue;        else        {            for(int j=i*2;j<maxn;j+=i)                su[j]=1;        }    }}bool judge(int x){    int i;    for(i=a;i<=b-x+1;i++)    {        if(sum[i+x-1]-sum[i-1]<k)            return 0;    }    return 1;}int main(){    Prime();    sum[1]=0;    for(int i=2;i<maxn;i++)    {        sum[i]=sum[i-1];        if(!su[i])            sum[i]++;    }    scanf("%d%d%d",&a,&b,&k);    int l=0,r=b-a+1,mid,ans=-1;    while(l<=r)    {        mid=(l+r)/2;        if(judge(mid))        {            ans=mid;            r=mid-1;        }        else            l=mid+1;    }    printf("%d\n",ans);    return 0;}

2016/12/10

再次写这道题，先是题意又理解错了，x(a ≤ x ≤ b - l + 1)，在x, x + 1, ..., x + l - 1上至少有k个素数

写成r=mid;和l=mid就TE！！！改成r=mid-1;和l=mid+1;就AC！！！！

代码同上（因为是又看了以前的才发现这次的错误呜）

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