hdu 4715 Difference Between Primes (打表 枚举)
Difference Between Primes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 528 Accepted Submission(s): 150
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
思路:
打个素数表,然后枚举b,判断a就够了。
ps:汗,开始以为xat为正数,一直WA,原来题目说的是绝对值。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #define maxn 10000005 using namespace std; int n,m,cxx,ans; bool vis[maxn]; int prime[664580]; void sieve(int nn) { int i,j,mm; mm=int(sqrt(nn+0.5)); memset(vis,0,sizeof(vis)); for(i=2;i<=mm;i++) { if(!vis[i]) { for(j=i*i;j<=nn;j+=i) { vis[j]=1; } } } } int get_prime(int nn) { int i,c=0; sieve(nn); for(i=2;i<=nn;i++) { if(!vis[i]) prime[c++]=i; } return c; } bool isprime(int x) { int i,j,t; t=sqrt(x+0.5); for(i=2;i<=t;i++) { if(x%i==0) return false ; } return true ; } int main() { int i,j,t,a,b,flag,sgn; cxx=get_prime(10000000); // 664579 scanf("%d",&t); while(t--) { scanf("%d",&n); flag=0; if(n==0) { printf("2 2\n"); continue ; } if(n>0) sgn=1; else sgn=0,n=-n; for(i=0;i<cxx;i++) { b=prime[i]; a=b+n; if(isprime(a)) { flag=1; if(sgn) printf("%d %d\n",a,b); else printf("%d %d\n",b,a); break ; } } if(!flag) printf("FAIL\n"); } return 0; }