codeforces 237C.Primes on Interval

C. Primes on Interval

time limit per test
2 seconds

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.

这题做得我好蛋疼。本来用的方法是用一数组存1至t的素数个数至一个数组中p[t].然后求使p[x+l-1]-p[x]>=k的最大t。本来这个方法是对的，只要检验每个x是否满足即可。但我那时想，x可能要跑10^6，若再让l跑10^6不就会超时吗Orz..一我没注意看有2s,二我把二分用在了别的地方，不是判断l的大小上。所以我立即换了个方法。

因为我们要找k个素数，所以我就分别讨论原数组的素数总和小于k个，等于k个，大于k个的情形。可惜，我错在了忘记判断该数是否会超界。。

现在终于AC了。

#include<stdio.h>
#include<string.h>
int p[1000001],cun[1000009],tot;
void biao(int n)
{
memset(p,0,sizeof(p));
p[1]=1;p[0]=1;tot=0;
for(int i=2;i<n;i++)
{
if(!p[i])
{
cun[tot++]=i;
for(int j=i+i;j<n;j+=i)
p[j]=1;
}
}
}

int main()
{
int T,i,k,j,a,b,xiao,da;
biao(1000001);
int front=0,rear=tot-1,mid;
while(~scanf("%d%d%d",&a,&b,&k))
{
{
front=0;
rear=tot-1;
int count,ans;
while(front<rear)
{
mid=front+(rear-front)/2;
if(cun[mid]>=a)rear=mid;
else front=mid+1;
}
front=rear;
count=0;
while(cun[rear]<=b)
{
count++;
if(count<=k){rear++;if(rear==tot)break;}
else break;;
}
if(count==k+1)
{
ans=cun[rear-1]-a+1;
for(;;)
{
a=cun[front];
if(ans<cun[rear]-a)ans=cun[rear]-a;
rear++;front++;
if(cun[rear]-1>=b||rear==tot)break;	//我就是忘记判断rear==tot?

}
if(ans<b-cun[front]+1)ans=b-cun[front]+1;
printf("%d\n",ans);
}
else if(count==k)
if(cun[rear-1]-a+1>b-cun[front]+1)
printf("%d\n",cun[rear-1]-a+1);
else printf("%d\n",b-cun[front]+1);
else printf("-1\n");
}
}
return 0;
}