LightOJ - 1259 Goldbach`s Conjecture【素筛&小细节】

LightOJ - 1259 Goldbach`s Conjectur.

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, …

题意：

给你一个n，满足哥德巴赫猜想的组数 (n=a+b，(a,b)均为质数)(n=a+b，(a,b)均为质数).

分析：

一个素筛+打表就好，但是内存有限制，小技巧点：筛素数时用bool类型定义数组.

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;

const int MAXN = 1e7 + 5;
bool a[MAXN];
int b[2000000];
int p = 0;

void init() {
a[0] = a[1] = 1;
for(int i = 2; i * i < MAXN; ++i) {
if(a[i]) continue;
for(int j = i * i; j < MAXN; j += i) {
a[j] = 1;
}
}
for(int i = 0; i < MAXN; i++) {
if(!a[i]) {
b[p++] = i;
}
}
//printf("%d\n", p);
}

int main() {
init();
int T, n;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
int ans = 0;
for(int i = 0; i < p; i++) {
if(n / 2 < b[i]) break;
if(!a[n - b[i]]) ans++;
}
static int t = 1;
printf("Case %d: %d\n", t++, ans);
}

4000
return 0;
}