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 1000000long 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[n]);    return 0;}