費馬定理類的問題的終結版

Double Happiness

On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as:
t = a2 + b2, 
where a, b are arbitrary positive integers.
Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
Input
The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3*10^8).
Output
In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
Sample test(s)
input
3 5
output
1
input
6 66
output
7

題目大意：

　　　　給定區間[L,R]，求區間內知足條件的數的個數：

　　　　條件1）z是素數

　　　　條件2）z=x^2+y^2 x和y爲任意的正整數

結題思路：

　　　　其實就是求知足費馬定理的數的個數。

　　　　費馬定理：一個奇素數z能夠表示成z=x^2+y^2的形式,當且僅當z能夠表示成4*t+1的時候。（偶素數2=1^2+1^2）

　　　　LR的範圍是1到3*10^8用普通的篩選法求素數表，時間空間都會超。使用兩次篩選來求素數。

　　　　3*10^8的平方根小於17500，用17500之內的素數能夠篩出範圍內的全部素數。在經過判斷是否知足z=t%4+1來累加。

　　　　又因爲z的範圍過大，但對於惟一肯定的z來講，t也惟一肯定，故能夠用t做爲數組下標即(z-1)/4。數組大小就會小4倍左右。

　　　　具體細節看代碼備註。

Code：

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <string>#include <bitset>#define MAXN 17500using namespace std;bitset<MAXN>ISP; //相似於bool型，能夠存01bitset<300000000/4+10>result;int prime[MAXN];int is_prime()//先篩選1-17500的素數{    int p=0;    ISP[0]=ISP[1]=1;    for (int i=1; i<=MAXN; i++)        if (ISP[i]==0)        {            prime[p++]=i;//用prime素組將素數存起來留到後面篩選用。。            for (int j=i+i; j<=MAXN; j+=i)                ISP[j]=1;        }    return p-1;}int cnt(int L,int R){    int p=is_prime();    for (int j=0; j<=p; j++)    {        int x=L/prime[j];        if (x*prime[j]<L) x++;        if (x==1) x++;        for (int k=x*prime[j]; k<=R; k+=prime[j]) //經過素數表再來篩選[L,R]中的素數               if (k%4==1) //標記符合條件2且不符合條件1的數                result[(k-1)/4]=1;    }    int cnt=0;    for (int i=L;i<=R;i+=4)        if (!result[(i-1)/4]) cnt++;    return cnt;}int main(){    is_prime();    int L,R;    int ok=0;    cin>>L>>R;    if (L<=2&&R>=2) ok=1; //2做爲偶素數且符合條件 單獨判斷    while (L%4!=1||L==1) //將區間邊界縮小到符合 %4==1        L++;    while (R%4!=1)        R--;    cout<<cnt(L,R)+ok;    return 0;}