Light oj 1259 Goldbach's Conjecture

最新推荐文章于 2019-02-24 01:10:27 发布

原创最新推荐文章于 2019-02-24 01:10:27 发布 · 255 阅读

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本文探讨了哥德巴赫猜想，并提供了一种算法来验证该猜想对于小于等于10^7的偶数是否成立。通过优化算法避免了超时问题，展示了筛选非素数的重要性。

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F - Goldbach`s Conjecture

LightOJ - 1259

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

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, ...

这道题看着比较水，但是数据太大了，直接开大数据会超时，需要筛去一些无用的非素数

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
bool prime[10000010] = {1, 1, 0};
int main()
{
    for(int i = 2; i * i <= 10000010; i ++)
        if(!prime[i])
        for(int j = i + i; j <= 10000010; j += i)
        prime[j] = 1;
    int t;
    int k = 1;
    cin>>t;
    while(t--)
    {
        int n;
        int sum = 0;
        cin>>n;
        for(int i = 1; i <= n/2; i ++)
        if(!prime[i]&&!prime[n - i])sum ++;
        printf("Case %d: %d\n",k++,sum);
    }
    return 0;
}

毫无疑问直接tle

改进之后的

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
const int maxn = 10000010;
bool prime[maxn];
int cnt = 0;
int countt[maxn/10];//筛去一些非素数
int main()
{
    memset(prime, 1, sizeof(prime));
    prime[0] = prime[1] = 0;
    for(int i = 2; i < maxn; i ++)
        if(prime[i])
        {
            countt[cnt++] = i;//记录
            for(int j = i + i; j <= maxn; j += i)
            prime[j] = 0;
        }
    int t;
    int k = 1;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        int sum = 0;
        scanf("%d",&n);
        for(int i = 0; countt[i] <= n/2 && i < cnt; i ++)
        if(prime[n - countt[i]])sum ++;
        printf("Case %d: %d\n",k++,sum);
    }
    return 0;
}