【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; }
This passage is made by MashiroSky.