[交流] 欧拉工程，第二十一题，计算10000以下亲和数的和。 已有5人参与

31626
elapsed time=0.020000 seconds.

#include
#include
#include

#define TIMERSTART clock_t start_time,stop_time;double elapsed_time;start_time = clock();
#define TIMERSTOP stop_time = clock();elapsed_time=(double)(stop_time-start_time)/CLOCKS_PER_SEC;printf("elapsed time=%f seconds.\n",elapsed_time);


int sumdivisors(int n){
  int i,sum=1,sqrtn=sqrt(n);
  for(i=2;i     if(n%i==0)sum+=i+n/i;
  }
  if(sqrtn*sqrtn==n)sum-=sqrtn;
  return sum;
}

int euler21(int n){
  int i,sum=0,tmp;
  for(i=3;i     tmp=sumdivisors(i);
    if(tmp!=i&&tmp       sum+=i;
    }
  }
  
  return sum;
}


int main(void){
int i;

TIMERSTART;

printf("%d\n",euler21(10000));
  
TIMERSTOP;

  return 0;
}

Originally posted by huycwork at 2011-06-02 22:34:29:
C++代码：
[code]
#include <iostream>
enum {BUFSZ = 10000};

size_t eular21(){
        size_t buf[BUFSZ];
        memset(buf, 0, sizeof buf);
        for(size_t i = 1; i < BUFSZ; ++i){
                for(size_t j = ...

Originally posted by dubo at 2011-06-04 14:55:25:
// play_time_calu.cpp : 定义控制台应用程序的入口点。
//

//#include "stdafx.h"

#include <stdio.h>
#include <math.h>
#include <time.h>

#define TIMERSTART cl ...