【poj2478】 Farey Sequence

http://poj.org/problem?id=2478 (题目链接)

题意

  求分母小于等于n的真分数的个数。

Solution

  现在只能做做水题了,唉,思维僵化。

细节

  前缀和开LL

代码

// poj2478
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define MOD 10000
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
 
const int maxn=1000010;
int phi[maxn],vis[maxn],p[maxn],n;
LL s[maxn];

void calphi() {
	phi[1]=1;
	for (int i=2;i<=maxn;i++) {
		if (!vis[i]) {p[++p[0]]=i;phi[i]=i-1;}
		for (int j=1;j<=p[0];j++) {
			if (p[j]*i>maxn) break;
			vis[p[j]*i]=1;
			if (i%p[j]==0) {phi[i*p[j]]=phi[i]*p[j];break;}
			else phi[i*p[j]]=phi[i]*phi[p[j]];
		}
	}
	for (int i=2;i<=maxn;i++) s[i]=s[i-1]+phi[i];
}
int main() {
	calphi();
	while (scanf("%d",&n)!=EOF && n) printf("%lld\n",s[n]);
    return 0;
}

  

posted @ 2016-12-13 10:37  MashiroSky  阅读(218)  评论(0编辑  收藏  举报