| Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 |
Output for Sample Input |
2 6 4 |
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, ...
题目大意：
首先给定T组数据，让你求的是，一个>=4的偶数拆分成两个素数之和的形式能够有几对。
样例解释：
当 n = 6时，6 = 3 + 3只有一对，
当 n=10时，10 = 5 + 5 = 3 + 7（7+3）有两对所以输出3
解题思路：
就是一个素数筛，没什么说的感觉（我还错了几次），当时也不知道是怎么回事，有可能是编译器坏了，明明一样的代码，在自己这就是不能过，但是交上去就能过，真是神了，需要注意的是素数筛的时候要记录素数的值，在主函数中用到，不能直接 n/2算，会超时，最后因为我们算的是对数所以要除以2，然后再判断 n/2是不是素数，如果是+1
不是不用变，上代码：
My Code：
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAXN = 1e7+5;
const int M = 1e6+5;
typedef long long LL;
int p[M];
bool prime[MAXN];
int cnt = 0;
void isprime()
{
cnt = 0;
memset(prime, true, sizeof(prime));
prime[0] = prime[1] = false;
for(LL i=2; i<MAXN; i++)
{
if(prime[i])
{
p[cnt++] = i;
for(LL j=i*i; j<MAXN; j+=i)
prime[j] = false;
}
}
}
int main()
{
isprime();
int T,n;
scanf("%d",&T);
for(int cas=1; cas<=T; cas++)
{
scanf("%d",&n);
int sum = 0;
for(int i=0; i<cnt&&p[i]<=n; i++)
if(prime[n-p[i]])
sum++;
sum >>= 1;
if(prime[n/2])
sum++;
printf("Case %d: %d\n",cas,sum);
}
return 0;
}
