标签:des style blog http color io os for sp
Constraints
1 <= T <= 300000
1 <= n <= 1000000
这道题将lcm转化为gcd并按照相同gcd分为一组的思路进行,巧妙地将题目转化为求小于等于n且与n互质数的和,而这个值时n*phi(n)/2
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 1000010 typedef long long qword; //segma(i*n/gcd(i,n)) //=n*segma(h(n/k)/k) //h(a)表示与a互质数的和 bool pflag[MAXN]; int prime[MAXN],topp=-1; int phi[MAXN]; void init() { int i,j; int x,y; for (i=2;i<MAXN;i++) { if (!pflag[i]) { prime[++topp]=i; phi[i]=i-1; } for (j=0;j<=topp && i*prime[j]<MAXN ;j++) { pflag[i*prime[j]]=true; phi[i*prime[j]]=phi[i]*phi[prime[j]]; if (i%prime[j]==0) { x=i;y=prime[j]; while (x%prime[j]==0) { x/=prime[j]; y*=prime[j]; } if (x==1) { phi[i*prime[j]]=i*(prime[j]-1); }else { phi[i*prime[j]]=phi[x]*phi[y]; } continue; } } } } qword h(int x) { if (x==1)return 1; return (qword)x*phi[x]/2; } int main() { freopen("input.txt","r",stdin); //freopen("output.txt","w",stdout); int nn; int n,i; scanf("%d",&nn); init(); while (nn--) { scanf("%d",&n); qword ans=0; for (i=1;i*i<n;i++) { if (n%i!=0)continue; ans+=(qword)n*h(n/i); ans+=(qword)n*h(i); } if (i*i==n) ans+=(qword)n*h(i); printf("%lld\n",ans); } }
bzoj 2226: [Spoj 5971] LCMSum 数论
标签:des style blog http color io os for sp
原文地址:http://www.cnblogs.com/mhy12345/p/4044668.html