POJ-2478-Farey Sequence-递推求欧拉函数

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

提示：这是一道递推求欧拉函数的模板题

递推求欧拉函数的算法为

for(int a = 1; a <= maxn; a ++)
i[a]=a;
for(int a = 2; a<= maxn; a +=2)
i[a]/=2;
for(int a = 3; a<=maxn; a +=2)
{
if(i[a]==a)
{
for(int b =a; b <= maxn; b +=a)
i[b]=i[b]/a*(a-1);
}
}

那么题解的代码如下

#include <cstdio>
#define maxn 1000000
long long i[1000005];
int main()
{
for(int a = 1; a <= maxn; a ++)
i[a]=a;
for(int a = 2; a<= maxn; a +=2)
i[a]/=2;
for(int a = 3; a<=maxn; a +=2)
{
if(i[a]==a)
{
for(int b =a; b <= maxn; b +=a)
i[b]=i[b]/a*(a-1);
}
}
for(int a = 3; a <= maxn; a ++)
i[a]+=i[a-1];
int n;
while(~scanf("%d",&n)&&n)
printf("%lld\n",i
);
return 0;
}