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 lintegers 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
給出a，b即一個區間，求一個最小的len，使得對所有的x，x,...,x+len-1這個區間內至少k個素數。
其中要滿足
a<=x<=b
a<=x-len+1<=b
二分len，利用前綴和優化。
#define ll long long
const ll mod=1e9+7;
ll n,m;
const ll N = 2000000+999;
ll prime[N] = {0},num_prime = 0; //prime存放着小於N的素數
int isNotPrime[N] = {1, 1}; // isNotPrime[i]如果i不是素數，則爲1
int Prime()
{
for(ll i = 2 ; i < N ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
//無論是否爲素數都會下來
for(ll j = 0 ; j < num_prime && i * prime[j] < N ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j] ) ) //遇到i最小的素數因子
//關鍵處1
break;
}
}
return 0;
}
預處理
Prime();
for(int i=1; i<=1000006; i++)
{
if(!isNotPrime[i])
{
s[i]=s[i-1]+1;
}
else
{
s[i]=s[i-1];
}
}
二分
二分寫法1 這個版本是求出符合答案最小的那個！
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll n,m;
const ll N = 2000000+999;
ll prime[N] = {0},num_prime = 0; //prime存放着小於N的素數
int isNotPrime[N] = {1, 1}; // isNotPrime[i]如果i不是素數，則爲1
int Prime()
{
for(ll i = 2 ; i < N ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
//無論是否爲素數都會下來
for(ll j = 0 ; j < num_prime && i * prime[j] < N ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j] ) ) //遇到i最小的素數因子
//關鍵處1
break;
}
}
return 0;
}
int s[N];
int a,b,k;
bool check(int len)
{
for(int i=a; i<=b-len+1; i++)
{
if(s[i+len-1]-s[i-1]<k)
return false;
}
return true;
}
int main()
{
Prime();
for(int i=1; i<=1000006; i++)
{
if(!isNotPrime[i])
{
s[i]=s[i-1]+1;
}
else
{
s[i]=s[i-1];
}
}
scanf("%d %d %d",&a,&b,&k);
int l=1;
int r=b-a+1;
int ans=999999;
while(l<r)
{
int mid=(l+r)/2;//長度
if(check(mid))
{
r=mid;
}
else
{
l=mid+1;
}
}
printf("%d\n",check(l)?l:-1);
}
要是求符合題意最大的 （>=x最大的一個數）：
while(l<r)
{
int mid=(l+r+1)/2;//長度
if(a[mid]<=x)
{
l=mid;
}
else
{
r=mid-1;
}
}
return a[l];
二分寫法2
這個方法要確保題目一定有答案的
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll n,m;
const ll N = 2000000+999;
ll prime[N] = {0},num_prime = 0; //prime存放着小於N的素數
int isNotPrime[N] = {1, 1}; // isNotPrime[i]如果i不是素數，則爲1
int Prime()
{
for(ll i = 2 ; i < N ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
//無論是否爲素數都會下來
for(ll j = 0 ; j < num_prime && i * prime[j] < N ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j] ) ) //遇到i最小的素數因子
//關鍵處1
break;
}
}
return 0;
}
int s[N];
int a,b,k;
bool check(int len)
{
for(int i=a; i<=b-len+1; i++)
{
if(s[i+len-1]-s[i-1]<k)
return false;
}
return true;
}
int main()
{
Prime();
for(int i=1; i<=1000006; i++)
{
if(!isNotPrime[i])
{
s[i]=s[i-1]+1;
}
else
{
s[i]=s[i-1];
}
}
scanf("%d %d %d",&a,&b,&k);
int l=1;
int r=b-a+1;
int ans=9999999;
while(l<=r)
{
int mid=(l+r)>>1;//長度
if(check(mid))
{
r=mid-1;
ans=min(ans,mid);
}
else
{
l=mid+1;
}
}
printf("%d\n",ans==9999999?-1:ans);
}