Farey Sequence(欧拉函数板子题)
题目链接:http://poj.org/problem?id=2478
Farey Sequence
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 19736 | Accepted: 7962 |
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.
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
Source
POJ Contest,Author:Mathematica@ZSU
题目大意:很容易可以发现是求2-n的所有数的欧拉函数值之和
看代码:
/** 有三条特性 若a为质数 phi[a]=a-1 若a为质数,b%a==0 phi[a*b]=phi[b]*a; 若a b 互质 phi[a*b]=phi[a]*phi[b](当a为质数 如果b%a!=0) */ #include<iostream> #include<cstdio> using namespace std; typedef long long LL; const int maxn=1e6+50; int phi[maxn],prime[maxn],p[maxn];//phi[i]代表i的欧拉函数值 prime[i]=0代表是素数 1代表不是素数 p存储素数 void make() { phi[1]=1;//特例 int num=0; for(int i=2;i<=maxn;i++) { if(!prime[i])//是素数 { p[num++]=i;// phi[i]=i-1;//素数的欧拉函数值就是它的值减1 } for(int j=0;j<num&&p[j]*i<maxn;j++)//用当前已经得到的素数筛去p[j]*i { prime[p[j]*i]=1;//可以确定p[j]*i不是质数 if(i%p[j]==0)//第二条特性 { phi[p[j]*i]=phi[i]*p[j]; break;//欧拉筛的核心语句 保证每个数只会被自己最小的质因子筛掉一次 } else phi[p[j]*i]=phi[i]*phi[p[j]]; } } return ; } int main() { make(); int n; while(scanf("%d",&n)!=EOF) { if(n==0) break; LL sum=0; for(int i=2;i<=n;i++) sum+=phi[i]; printf("%lld\n",sum); } // for(int i=1;i<=100;i++) cout<<phi[i]<<" "; return 0; }
当初的梦想实现了吗,事到如今只好放弃吗~