C. Primes on Interval
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
許久不大量做題，感覺手生了，一些細節上的錯誤硬是發現不了。
本題考查二分，外加素數打表的基本套路。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1000010;
bool vis[MAXN];
int sum[MAXN];
void getvis() {
vis[1]=true;
for(int i=2; i<MAXN; i++) {
if(vis[i])
continue;
for(int j=i+i; j<MAXN; j+=i) {
vis[j]=true; //坑的很啊，j寫成i硬是沒發現
}
}
sum[0] = 0;
for(int i = 1; i < MAXN; i++) {
sum[i] = sum[i-1] + (vis[i] == false);
}
}
bool judge(int o,int a,int b,int k) {
for(int i=a; i<=b-o+1; i++) {
if(sum[i+o-1]-sum[i-1]<k)
return false;
}
return true;
}
int main() {
getvis();
int a,b,k;
while(scanf("%d%d%d",&a,&b,&k)!=EOF) {
int l=1,r=b-a+1;
int ans=-1;
while(r>=l) {
int mid=(l+r)>>1;
if(judge(mid,a,b,k)) {
ans=mid;
r=mid-1;
} else {
l=mid+1;
}
}
printf("%d\n",ans);
}
return 0;
}