1. 程式人生 > >【題解】LightOJ1259 Goldbach`s Conjecture 線性篩

【題解】LightOJ1259 Goldbach`s Conjecture 線性篩

題目連結

Description

Goldbach’s conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:

Every even integer, greater than 2, can be expressed as the sum of two primes [1].

Now your task is to check whether this conjecture holds for integers up to 107.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case starts with a line containing an integer n (4 ≤ n ≤ 107, n is even).

Output

For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of (a, b) where

1) Both a and b are prime

2) a + b = n

3) a ≤ b

Sample Input

2

6

4

Sample Output

Case 1: 1

Case 2: 1

Note

1.An integer is said to be prime, if it is divisible by exactly two different integers. First few primes are 2, 3, 5, 7, 11, 13, …

我居然能想到n²的列舉……真是為我智商捉急

#include<cstdio>
const int N=1e7+10; int t,prime[N/2],p,n,ca; bool iscomp[N]; void primetable() { for(int i=2;i<N;i++) { if(!iscomp[i])prime[p++]=i; for(int j=0;j<p&&prime[j]*i<N;j++) { iscomp[i*prime[j]]=1; if(i%prime[j]==0)break; } } } int main() { //freopen("in.txt","r",stdin); primetable(); scanf("%d",&t); while(t--) { scanf("%d",&n); int cnt=0; for(int i=0;prime[i]<=n/2;i++) if(!iscomp[n-prime[i]])cnt++; printf("Case %d: %d\n",++ca,cnt); } return 0; }

總結

線性篩出素數之後直接暴力就好