UVA 11426 gcd求和
UVA 11426 gcd求和
Description
Problem J
GCD Extreme (II)
Input: Standard Input
Output: Standard Output
Given the value of N, you will have to find the value of G. The definition of G is given below:
Here GCD(i,j) means the greatest common divisor of integer i and integer j.
For those who have trouble understanding summation notation, the meaning of G is given in the following code:
G=0; for(i=1;i<N;i++) for(j=i+1;j<=N;j++) { G+=gcd(i,j); } /*Here gcd() is a function that finds the greatest common divisor of the two input numbers*/ |
Input
The input file contains at most 100 lines of inputs. Each line contains an integer N (1<N<4000001). The meaning of N is given in the problem statement. Input is terminated by a line containing a single zero.
Output
For each line of input produce one line of output. This line contains the value of G for the corresponding N. The value of G will fit in a 64-bit signed integer.
Sample Input Output for Sample Input
10 100 200000 0
|
67 13015 143295493160
|
Problemsetter: Shahriar Manzoor
Special Thanks: SyedMonowarHossain
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> #define ll long long #define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) #define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t)) #define PII pair<int,int> #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&(x)) #define RLL(x) scanf("%lld",&(x)) #define RI64(x) scanf("%I64d",&(x)) #define DRI(x) int x;scanf("%d",&(x)) #define DRLL(x) ll x;scanf("%lld",&(x)) #define DRI64(x) llx;scanf("%I64d",&(x)) #define MS0(a) memset((a),0,sizeof((a))) #define MS1(a) memset((a),0,sizeof((a))) #define MS(a,b) memset((a),(b),sizeof((a))) using namespace std; const int maxn=4000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); ll n; int euler[maxn]; ll F[maxn],f[maxn]; void getEuler() { MS0(euler); euler[1]=1; REP(i,2,maxn-1){ if(!euler[i]){ REPP(j,i,maxn-1,i){ if(!euler[j]) euler[j]=j; euler[j]=euler[j]/i*(i-1); } } } } int main() { getEuler(); MS0(f); REP(i,1,maxn-1){ REPP(j,i*2,maxn-1,i) f[j]+=i*euler[j/i]; } MS0(F); REP(i,1,maxn-1) F[i]=F[i-1]+f[i]; while(cin>>n,n){ cout<<F[n]<<endl; } return 0; }