【codeforces237C】Primes on Interval

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

许久不大量做题，感觉手生了，一些细节上的错误硬是发现不了。

本题考查二分，外加素数打表的基本套路。

#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写成i硬是没发现}}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;}