LightOJ 1259 - Goldbach`s Conjecture （分解偶数为两个素数之和）

1259 - Goldbach`s Conjecture

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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

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 10000010
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
bool v[MAXN];
int cnt;
int prime[666666];
void db()
{
cnt=0;v[1]=true;
for(int i=2;i<=10000000;i++)
{
if(!v[i])
{
prime[cnt++]=i;
for(int j=i*2;j<=10000000;j+=i)
v[j]=true;
}
}
}
int main()
{
db();
int t,cas=0,i;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int ans=0;
for(i=0;i<cnt;i++)
{
if(prime[i]>n/2+1)
break;
if(!v[n-prime[i]]&&n>=prime[i]*2)
ans++;
}
printf("Case %d: ",++cas);
printf("%d\n",ans);
}
return 0;
}