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poj 2478 欧拉函数筛选

2016-08-13 10:06 211 查看
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Farey Sequence

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14952 Accepted: 5927
Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 

F2 = {1/2} 

F3 = {1/3, 1/2, 2/3} 

F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 

F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5} 

You task is to calculate the number of terms in the Farey sequence Fn.
Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input.
Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input
2
3
4
5
0

Sample Output
1
3
5
9

在自己电脑上面跑溢出，debug时候卡的编译器崩溃

然后今天上午就重启了好几次

在程序中利用欧拉函数如下性质，可以快速求出欧拉函数的值 ( P为N的质因子 )

　　　　若(N%P==0 && (N/P)%P==0) 则有:E(N)=E(N/P)*P;

 

　　　　若(N%P==0 && (N/P)%P!=0) 则有:E(N)=E(N/P)*(P-1);

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define maxn 1000010
long long eu[maxn];

void init()
{
memset(eu,0,sizeof(eu));
for(int i=2;i<=maxn;i++){
if(!eu[i]){
for(int j=i;j<=maxn;j+=i){
if(!eu[j])
eu[j]=j;
eu[j]=eu[j]/i*(i-1);
}
}
}
}

int main()
{
int n;
init();
//freopen("in.txt","r",stdin);
while(scanf("%d",&n),n)
{
long long ans=0;
for(int i=1;i<=n;i++)
ans+=eu[i];
printf("%lld\n",ans);
}
return 0;
}
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