Fermat’s Chirstmas Theorem
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描写叙述
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
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.
演示样例输入
10 20
11 19
100 1000
-1 -1
演示样例输出
10 20 4 2
11 19 4 2
100 1000 143 69
果然蛋疼的一道题。题意说的非常清楚，就用素数筛暴力就能够了，有一个坑就是比方范围是 1-2 这时1也是符合条件的，由于 1==4*0+1且1==0*0+1*1(尽管1不是素数。但为什么会有有这样的数据？)
<pre name="code" class="html">#include <iostream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <vector>
#include <string>
#include <map>
#include <queue>
using namespace std;
const int maxn= 1000010;
int num=0;
int vis[maxn],prime[maxn];
void init_prime()
{
memset(vis,1,sizeof(vis));
vis[0]=0;vis[1]=0;
for(int i=0;i<=maxn;i++)
{
if(vis[i])
{
prime[++num]=i;
for(int j=1;j*i<=maxn;j++)
vis[j*i]=0;
}
}
}
int main()
{
int L,U,i;
init_prime();
while(scanf("%d%d",&L,&U)!=EOF)
{
int cnt1=0,cnt2=0;
if(L==-1&&U==-1) break;
for(i=0;i<=num;i++)
{
if(prime[i]&&prime[i]>=L&&prime[i]<=U)
{
if((prime[i]-1)%4==0)
cnt2++;
cnt1++;
}
}
if(L<=2&&U>=2)
cnt2++;
printf("%d %d %d %d\n",L,U,cnt1,cnt2);
}
return 0;
}