Goldbach`s Conjecture LightOJ

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
Hint
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, …

这道题显示超内存，后又是超内存。。。
1，首先找出素数，并存放在数组prime[] 中，（如果直接搜的话会超时）
2，标记是否为素数的数组用bool类型，否则超内存，prime[] 数组应比mark[] 数组小，否则也超内存。
3，这道题不难，细节多了也就难了，平时还是要多注意细节问题。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<queue>#include<algorithm>#define ll long long#define inf 0x3f3f3f3f#define maxn 10000007  // 1e7+7using namespace std;bool mark[maxn];int prime[maxn/10];int cnt = 0;void find_prime(){    mark[0] = true;    mark[1] = true;    for(int i = 2; i < maxn; i ++)      {        if(mark[i] == false)        {            prime[cnt++] = i;            for(int j = i + i; j < maxn; j += i)            {                mark[j] = true;            }        }    }}int main(){    int t;    int n;    find_prime();//  printf("cnt = %d\n",cnt);    scanf("%d",&t);    for(int i = 1; i <= t; i ++)    {        scanf("%d",&n);        int tmp = n / 2;        int cnt = 0;        for(int j = 0; prime[j] <= tmp; j ++)            if(mark[ prime[j] ] == 0 && mark[ n - prime[j] ] == 0)                cnt ++;        printf("Case %d: %d\n",i,cnt);    }    return 0;}