HDU 4715:Difference Between Primes
2016-05-01 20:05
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Difference Between Primes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3339 Accepted Submission(s): 953
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
迷失在幽谷中的鸟儿,独自飞翔在这偌大的天地间,却不知自己该飞往何方……
#include<iostream> using namespace std; const long N = 1000005; long prime ,num_prime; int isNotPrime = {1, 1}; void getprime() { for(long i = 2 ; i < N ; i ++) { if(!isNotPrime[i])prime[num_prime ++]=i; for(long j = 0 ; j < num_prime && i * prime[j] < N ; j ++) { isNotPrime[i * prime[j]] = 1; if(!(i%prime[j]))break; } } } int main() { getprime(); int T,n,s,a; cin>>T; while(T--) { cin>>n; a=n>0?n:-n; s=0; for(int i=0; i<100000; i++) if(!isNotPrime[prime[i]+a]) { if(n>0) cout<<prime[i]+a<<" "<<prime[i]<<endl; else cout<<prime[i]<<" "<<prime[i]+a<<endl; s=1; break; } if(s==0)cout<<"FAIL"<<endl; } return 0; }
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