Difference Between Primes
Problem Description
All you know Goldbach conjecture.That is to say, Every even integer greater than 2 can be expressed as the sum of two primes. Today, skywind present a new conjecture: every even integer can be expressed as the difference of two primes. To validate this conjecture, you are asked to write a program.
Input
The first line of input is a number nidentified the count of test cases(n<10^5). There is a even number xat the next nlines. The absolute value of xis not greater than 10^6.
Output
For each number xtested, outputstwo primes aand bat one line separatedwith one space where a-b=x. If more than one group can meet it, output the minimum group. If no primes can satisfy it, output ‘FAIL’.
Sample Input
3
6
10
20
Sample Output
11 5
13 3
23 3
Source
2013 ACM/ICPC Asia Regional Online —— Warmup
#include<bits/stdc++.h>
using namespace std;
const int maxn=1000005;
int vis[2*maxn];
int prime[maxn];
int total;
void makeprime()
{
total=0;
memset(vis,0,sizeof(vis));
prime[total++]=2;
vis[2]=1;
int flag;
for(int i=3; i<=maxn; i+=2)
{
flag=1;
for(int j=0; prime[j]*prime[j]<=i; j++)
{
if(i%prime[j]==0)
{
flag=0;
break;
}
}
if(flag)
{
prime[total++]=i;
vis[i]=1;
}
}
}
int main()
{
makeprime();
int t,n,i;
cin>>t;
while(t--)
{
cin>>n;
for(i=0; i<=total; i++)
{
if(vis[prime[i]+abs(n)])
break;
}
if(n>0)printf("%d %d\n",prime[i]+abs(n),prime[i]);
else printf("%d %d\n",prime[i],prime[i]+abs(n));
}
}
思路:
可能是水题吧,我太菜感觉还是有点难度。
先素数打表,然后把素数标记,然后从小到大开始遍历,如果是素数就停下来,这样满足最小的两个差。