Codeforces 237C Primes on Interval（素数统计）

http://codeforces.com/problemset/problem/237/C

C. Primes on Interval

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard 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

题意：

在给定的区间  [a, b] 之内，问是否存在一个最小的整数 l ，使得任意一个在区间内（a ≤ x ≤ b - l + 1）的
x ，至少在 x, x + 1, ..., x + l - 1 有 k 个素数。

思路：

统计每个数字之前的素数个数，二分即可。

就这题，四级怎么办？！



AC CODE：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
using namespace std;
const int MYDD = 1103+1e6;

bool isprime[MYDD];
int cntisp[MYDD];
void init() {
memset(isprime, 1, sizeof(isprime));
isprime[0] = 0, isprime[1] = 0;
for(int j = 2; j < MYDD; j++) {
if(isprime[j]) {
for(int k = 2*j; k < MYDD; k = k+j)
isprime[k] = 0;
}
}

cntisp[0] = 0, cntisp[1] = 0;//当前数字之前素数个数统计
for(int j = 2; j < MYDD; j++) {
cntisp[j] = cntisp[j-1];
if(isprime[j]) cntisp[j]++;
}
}

int HardBoy {
init();
int a, b, k;
scanf("%d %d %d", &a, &b, &k);

int l = k, r = b-a+1;
while(l <= r) {
int j, mid = (l+r)/2;
for(j = a; j+mid-1 <= b; j++) {
if(cntisp[j+mid-1] - cntisp[j-1] < k) break;
}
if(j+mid-1 <= b) l = mid+1;
else r = mid-1;
}

if(l > b-a+1) puts("-1");
else
printf("%d\n", l);

ForMyLove
}