X - Farey Sequence POJ - 2478

最新推荐文章于 2019-09-09 20:21:46 发布

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最新推荐文章于 2019-09-09 20:21:46 发布

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本文链接：https://blog.youkuaiyun.com/lihuanz/article/details/82319157

本文介绍了一种使用欧拉函数来高效计算Farey序列中项数的方法，并提供了一个C++实现示例，该方法可以快速计算任意整数n对应的Farey序列中的项数。

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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 <= 10 6). 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

Sample Output

题解：这一题就是欧拉函数的模拟题，做题的时候，没有考虑数值的大小，就做错了。

#include<stdio.h>
#include<string.h>
__int64 e[1000005],ans[1000005];
void fun()  
{
    int i,j;
    for(i=2; i<1000005; i++)
        if(!e[i])
            for(j=i; j<1000005; j+=i)
            {
                if(!e[j])
                    e[j]=j;
                e[j]=e[j]-e[j]/i;
            }
    for(i=1; i<1000005; i++)
        ans[i]=ans[i-1]+e[i];
}
int main()
{
    int n;
    fun();
    while(~scanf("%d",&n)&&n)
        printf("%I64d\n",ans[n]);
}