Language:
Farey Sequence
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;
}