Fermat’s Chirstmas Theorem

最新推荐文章于 2014-09-02 09:04:46 发布

ACBOY、

最新推荐文章于 2014-09-02 09:04:46 发布

阅读量737

点赞数

CC 4.0 BY-SA版权

分类专栏：数学

本文链接：https://blog.youkuaiyun.com/crescent__moon/article/details/8691320

数学专栏收录该内容

153 篇文章

订阅专栏

本文介绍了费马圣诞节定理，即一个奇质数可以表示为两个平方数之和当且仅当它可以表示为4的倍数加1。通过提供示例和代码实现，详细解释了如何计算给定区间内能被表示为平方和的素数数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

点击打开链接

Fermat’s Chirstmas Theorem

Time Limit: 1000MS Memory limit: 65536K

题目描述

In a letter dated December 25, 1640; the great mathematician Pierre de Fermat wrote to Marin Mersenne that he just proved that an odd prime p is expressible as p = a2 + b2 if and only if p is expressible as p = 4c + 1. As usual, Fermat didn’t include the proof, and as far as we know, never

wrote it down. It wasn’t until 100 years later that no one other than Euler proved this theorem.

To illustrate, each of the following primes can be expressed as the sum of two squares:

5 = 2 ² + 1 ²

13 = 3 ² + 2 ²

17 = 4 ² + 1 ²

41 = 5 ² + 4 ²

Whereas the primes 11, 19, 23, and 31 cannot be expressed as a sum of two squares. Write a program to count the number of primes that can be expressed as sum of squares within a given interval.

输入

Your program will be tested on one or more test cases. Each test case is specified on a separate input line that specifies two integers L, U where L ≤ U < 1, 000, 000

The last line of the input file includes a dummy test case with both L = U = −1.

输出

L U x y

where L and U are as specified in the input. x is the total number of primes within the interval [L, U ] (inclusive,) and y is the total number of primes (also within [L, U ]) that can be expressed as a sum of squares.

示例输入

示例输出

10 20 4 2
11 19 4 2
100 1000 143 69

借用了一套高效素数打表的方法。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
bool visit[1010000];
int prime[1000100];
int num = 0;
void init_prim(int n)
{
    memset(visit, true, sizeof(visit));
    for (int i = 2; i <= n; ++i)
    {
        if (visit[i] == true)
        {
            num++;
            prime[num] = i;
        }
        for (int j = 1; ((j <= num) && (i * prime[j] <= n));  ++j)
        {
            visit[i * prime[j]] = false;
            if (i % prime[j] == 0) break;
        }
    }
}
int main()
{
    init_prim(1000000);
    visit[0]=0;
    visit[1]=0;
    int l,p;
    int ans1,ans2;
    int x,y;
    while(scanf("%d%d",&l,&p))
    {
        if(l==-1&&p==-1)
            break;
        ans1=ans2=0;
        for(int i=0; i<=num; i++)
            if(prime[i]&&prime[i]>=l&&prime[i]<=p)
            {
                if((prime[i]-1)%4==0)
                    ans2++;
                ans1++;
            }
        if(l<=2&&p>=2)
            ans2++;
        printf("%d %d %d %d\n",l,p,ans1,ans2);
    }
    return 0;
}