UVA 11426 GCD - Extreme (II)（神TM GCD大法，欧拉函数）

Given the value of N, you will have to findthe value of G. The definition of G is given below:

Here GCD(i,j)means the greatest common divisor ofinteger i andinteger j.

 

For those who have trouble understandingsummation notation, the meaning of G is given in the followingcode:

G=0;

for(i=1;i<N;i++)

for(j=i+1;j<=N;j++)

{

    G+=gcd(i,j);

}

 

 

Input

The input file contains at most 100 lines ofinputs. Each line contains an integer N(1<N<4000001)

 

Output

For each line of input produce one line ofoutput. This line contains the value of G for the corresponding N.The value of G will fit in a 64-bit signed integer.

 

SampleInput 

10

100

200000

0

Output for SampleInput

67

13015

143295493160

题意：求对于每个i<=N,求f(i)=GCD(1,i)+GCD(2,i)+......GCD(i-1,i)并求前N项和。

对于一个N，设GCD(a,N)=b;则有N=b*x，a=b*y,GCD(a/b,N/b)=1即GCD(x,y)=1;

又依题意得a<N,则Y<X,则满足GCD(a,N)=b且a<N的a的个数等于GCD(y,x)=1且y<x的y的个数，即eluer(x)（小于x且与x互质的数的个数）

因此打个欧拉表再枚举N的约数b即可

代码：

#include<cstdio>#include<cstring>#include<string>#include<iostream>#include<sstream>#include<algorithm>#include<utility>#include<vector>#include<set>#include<map>#include<queue>#include<math.h>#include<iterator>#include<stack>using namespace std;typedef long long LL;typedef unsigned long long ULL;const double eps=1e-8,PI=3.1415926538;const LL MOD=1000000000+7;const LL MAXN=4000007;LL m[MAXN],phi[MAXN],p[MAXN],pt;void make(){    phi[1]=1;    LL N=MAXN;    LL k;    for(int i=2;i<N;i++)    {        if(!m[i])            p[pt++]=m[i]=i,phi[i]=i-1;        for(int j=0;j<pt&&(k=p[j]*i)<N;j++)        {            m[k]=p[j];            if(m[i]==p[j])            {                phi[k]=phi[i]*p[j];                break;            }            else                phi[k]=phi[i]*(p[j]-1);        }    }}LL gcd(LL a,LL b){    if(b==0)return a;    else return gcd(b,a%b);}LL S[MAXN];int main(){    make();    phi[1]=phi[0]=0;    memset(S,0,sizeof(S));    LL z=sqrt(MAXN);    for(int i=1;i<=z;i++)//枚举N的约数    {        for(int j=2;j*i<=MAXN-3;j++)//N=i*j        {            S[j*i]+=i*phi[j];            if(z<j)S[j*i]+=j*phi[i];//z<span style="font-family:Courier New;"><j时，i*j的约数j枚举不到</span>        }    }    LL N,G;    while(scanf("%lld",&N)!=-1&&N!=0)    {        G=0;        for(int i=2;i<=N;i++)        {            G+=S[i];        }        printf("%lld\n",G);    }    return 0;}