Fermat’s Chirstmas Theorem

                                                 Fermat’s Chirstmas Theorem

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 = 2² + 1²

13 = 3² + 2²

17 = 4² + 1²

41 = 5² + 4²

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

10 2011 19100 1000-1 -1

示例输出

10 20 4 211 19 4 2100 1000 143 69

题意为找所给区间中的素数以及可以表示成两个数平方和的素数个数

#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<queue>#include<set>#include<string>using namespace std;int Max=1000000;int prime[1000000];int flag[1000000];int num;void sushu(){    int i,j;    num=0;    memset(flag,0,sizeof(flag));    for(i=2;i<Max;i++)    {        if(flag[i]==0) prime[num++]=i;        for(j=0;j<num&&i*prime[j]<Max;j++)        {            flag[i*prime[j]]=1;            if(i%prime[j]==0) break;        }    }}int main(){    int l,u,x,y,i,j;    sushu();    while(scanf("%d%d",&l,&u)!=EOF)    {        x=0;y=0;        if(l==-1&&u==-1) break;        for(i=0;i<num;i++)        {            if(prime[i]>=l&&prime[i]<=u)            {              x++;              if(prime[i]%4==1)              y++;            }            if(prime[i]>u) break;        }        if(l<=2&&u>=2) y++;//注意：2既是素数也可以表示为两个数的平方和        printf("%d %d %d %d\n",l,u,x,y);    }}