LightOJ 1259

1259 - Goldbach`s Conjecture

   PDF (English)StatisticsForum

Time Limit: 2 second(s)Memory Limit: 32 MB

Goldbach's conjecture is one of the oldest unsolved problemsin 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 forintegers 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 casenumber and the number of ways you can express n as sum of two primes. Tobe more specific, we want to find the number of (a, b) where

1)      Both aand 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.      Aninteger is said to be prime, if it is divisible by exactly two differentintegers. First few primes are 2, 3, 5, 7, 11, 13, ...

题意：给你一个偶数n，问有多少种方案找到两个数a,b，且a+b=n,a<=b，a和b是素数

思路：直接打个表就行了，然后暴力验证，但是用int来标记数会超内存，所以说改成bool就好了

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>using namespace std;typedef long long ll;const int maxn = 1e7+5;bool prim[maxn];int primes[700000], p;void init(){    for(int i = 2; i < maxn; i++)        if(!prim[i])        {            primes[p++] = i;            for(int j = i+i; j < maxn; j += i)                prim[j] = true;        }}int main(){    init();    int n, T;    scanf("%d", &T);    for(int kase =1; kase <= T; kase++)    {        scanf("%d", &n);        int ans = 0;        for(int i = 0; primes[i] <= n/2; i++)        {            if(!prim[n-primes[i]])                ans++;        }        printf("Case %d: %d\n", kase, ans);    }    return 0;}