[交流] Euler 工程第十六题：2的1000次方的各项和 已有6人参与

1366
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
elapsed time=0.000000 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);

#define N 100

static int num[N];

void multiply(int n) {
  int i=0,tmp,residual=0;
  for(i=0;i     tmp=num[i]*n+residual;
    num[i]=tmp%10000;
    residual=(tmp-num[i])/10000;
  }
  return;
}

int euler16(int n){
  int i,sum=0;
  
  for(i=1;i     multiply(2);
  }
  
  for(i=0;i     sum+=(num[i]/1000+(num[i]%1000)/100+(num[i]%100)/10+num[i]%10);
  }
  
  return sum;
}


int main(void){
int i;

TIMERSTART;
num[0]=1;

printf("%d\n",euler16(1000));

for(i=N-1;i>=0;i--){
   printf("%04d",num[i]);
}
printf("\n");


TIMERSTOP;

  return 0;
}

#!/usr/bin/perl
use bigint;
$sum += $_ foreach(2**1000 =~ /(.)/g);
print "result:$sum\n";

Originally posted by huycwork at 2011-05-26 21:38:51:
来个Perl版：

CODE:

#!/usr/bin/perl
use bigint;
$sum += $_ foreach(2**1000 =~ /(.)/g);
print "result:$sum\n";

结果：1366

Originally posted by sudo at 2011-05-26 21:52:34:
咳，我突然想到一个好玩的东西...

对于一个数的大指数出来的各个位数，0-9的概率分布趋向于均匀分布~

那么我们可以做大致的估算：

2^1000的位数为

向下取整(1000 * log 2) + 1 = 302 位

对于 ...