POJ 2689 Prime Distance （埃氏筛 区间筛）

原题链接（英文题面）
中文题面
acwing上wa时会有错误数据，还是很喜欢的~
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Prime Distance

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Description

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).
Input

Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.
Output

For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.
Sample Input

2 17
14 17
Sample Output

2,3 are closest, 7,11 are most distant.
There are no adjacent primes.
Source

Waterloo local 1998.10.17

题解以后补吧
数组开小了段错误，开大了内存超限…
还有这是多组输入

#include <cstring>
#include <iostream>
#include <algorithm>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int N = 1e5;
int prim[N];
int judge[N];//0为素数，1为质数 
int cnt=0;
void init(){
	for(ll i=2;i<N;i++){
		if(!judge[i])
			prim[++cnt]=i;
			for(ll j=1;j<=cnt&&i*prim[j]<N;j++){
				judge[prim[j]*i]=1;
				if(i%prim[j]==0) break;
			}
	}
}
int mp[N];
int prim2[N];
int ans=0;
void qvjian(ll l,ll r){
	 ans=0;
	memset(prim2,0,sizeof prim2);
	memset(mp,0,sizeof mp);
	if(l==1) l=2;
	for(int i=1;i<=cnt&&prim[i]*prim[i]<=r;i++){
		//ll x=l/prim[i];
		//x*=prim[i];
		//if(l%prim[i]==0){
	//		x+=prim[i];
	//		if(l!=2) mp[0]=1;
	//	}  
		ll x=(l+prim[i]-1)/prim[i]*prim[i];
		if(x==prim[i]) x+=prim[i];
		//cout<<prim[i]<<" "<<x<<endl;
		
		for(ll j=x;j<=r;j+=prim[i]) 
			mp[j-l]=1;//合数 
		
	}
	for(int i=0;i<=r-l;i++) if(!mp[i]) prim2[++ans]=i+l;
}
int main(){
	init();
	//for(int i=1;i<10;i++) cout<<prim[i]<<endl;
//	cout<<endl;
	ll l,r;
	while(~scanf("%lld%lld",&l,&r)){
		qvjian(l,r);
		//for(int i=1;i<=ans;i++) cout<<prim2[i]<<endl;
		if(ans<2) cout<<"There are no adjacent primes."<<endl;
		else{
			int minn=1,maxx=1;
			for(int i=1;i<ans;i++){
				int t=prim2[i+1]-prim2[i];
				if(t<prim2[minn+1]-prim2[minn]) minn=i;
				if(t>prim2[maxx+1]-prim2[maxx]) maxx=i;
			}
			printf("%d,%d are closest, %d,%d are most distant.\n",prim2[minn],prim2[minn+1],prim2[maxx],prim2[maxx+1]);
		}
	}
	return 0;
}