| ²é¿´: 2282 | »Ø¸´: 14 | |||
| ±¾Ìû²úÉú 3 ¸ö ³ÌÐòÇ¿Ìû £¬µã»÷ÕâÀï½øÐв鿴 | |||
| µ±Ç°Ö»ÏÔʾÂú×ãÖ¸¶¨Ìõ¼þµÄ»ØÌû£¬µã»÷ÕâÀï²é¿´±¾»°ÌâµÄËùÓлØÌû | |||
huycwork½ð³æ (ÖøÃûдÊÖ)
|
[½»Á÷]
Å·À¹¤³Ì£¬µÚ¶þʮһÌ⣬¼ÆËã10000ÒÔÏÂÇ׺ÍÊýµÄºÍ¡£ ÒÑÓÐ5È˲ÎÓë
|
||
|
¿È¿È£¬·Å¼ÙÓë˯¾õ֮ǰÔÙ·¢Ò»Ì⣬EÎIJ»ºÃ£¬Öîλ¼ûÁ¹þ~ ¶¨Òåd(n)ÊÇnµÄËùÓÐÔ¼ÊýµÄºÍ¡£ Èç¹ûd(a) = b£¬d(b) = aÇÒÂú×ãa != b£¬Ôò˵aÓëbÊÇÒ»×éÇ׺ÍÊý¡£ ÀýÈ磬220µÄËùÓÐÔ¼ÊýÊÇ1, 2, 4, 5, 10, 11, 20, 22, 44, 55Óë110£¬Ôòd(220) = 284£»¶ø284µÄËùÓÐÔ¼ÊýÊÇ1, 2, 4, 71Óë142£¬Àۼӵãºd(284) = 220¡£ ÊÔ¼ÆËã10000ÒÔÏÂËùÓÐÇ׺ÍÊýÖ®ºÍ¡£ |
» ²ÂÄãϲ»¶
080200ѧ˶£¬»úе¹¤³Ìרҵ277·Ö£¬Çó´ø×ߣ¡
ÒѾÓÐ7È˻ظ´
377Çóµ÷¼Á
ÒѾÓÐ3È˻ظ´
295Çóµ÷¼Á
ÒѾÓÐ8È˻ظ´
086000ÉúÎïÓëÒ½Ò©298µ÷¼ÁÇóÖú
ÒѾÓÐ10È˻ظ´
0860 Çóµ÷¼Á Ò»Ö¾Ô¸¹ú¿Æ´ó 348 ·Ö
ÒѾÓÐ3È˻ظ´
²ÄÁϵ÷¼Á
ÒѾÓÐ4È˻ظ´
±¾¿Æ211ÉúÎïҽѧ¹¤³Ì085409Çóµ÷¼Á339·Ö
ÒѾÓÐ9È˻ظ´
296Çóµ÷¼Á
ÒѾÓÐ5È˻ظ´
385·Ö ÉúÎïѧ£¨071000£©Çóµ÷¼Á
ÒѾÓÐ11È˻ظ´
308Çóµ÷¼Á
ÒѾÓÐ3È˻ظ´
» ±¾Ö÷ÌâÏà¹Ø¼ÛÖµÌùÍÆ¼ö£¬¶ÔÄúͬÑùÓаïÖú:
Project Euler 50 Å·À¹¤³Ì 50 Ìâ
ÒѾÓÐ12È˻ظ´
fluentµÄÅ·À-À¸ñÀÊÈÕ·¨ÇóÖú
ÒѾÓÐ15È˻ظ´
Å·ÀÆø¹ÌÁ½ÏàÁ÷Ä£Äâ
ÒѾÓÐ5È˻ظ´
Å·ÀÄ£ÐÍ
ÒѾÓÐ5È˻ظ´
Project Euler 48 Å·À¹¤³Ì 48 Ìâ
ÒѾÓÐ30È˻ظ´
Project Euler 45 Å·À¹¤³Ì 45 Ìâ
ÒѾÓÐ7È˻ظ´
Euler Project Q17. Å·À¹¤³ÌµÚÊ®ÆßÌâ
ÒѾÓÐ4È˻ظ´
Euler Project Q13 Å·À¹¤³ÌµÚÊ®ÈýÌâ
ÒѾÓÐ20È˻ظ´
Euler Project Q12 Å·À¹¤³ÌµÚÊ®¶þÌâ
ÒѾÓÐ23È˻ظ´
Euler Project Q8. Å·À¹¤³ÌµÚ°ËÌâ
ÒѾÓÐ4È˻ظ´
Euler Project Q7. Å·À¹¤³ÌµÚÆßÌâ
ÒѾÓÐ14È˻ظ´
¡¾ÇóÖú¡¿Å·ÀÄ£ÐÍÖпÅÁ£Ïàgranular temperature¶¨Òå
ÒѾÓÐ6È˻ظ´
¡¾ÇóÖú¡¿fluentÄ£ÄâÆø¹ÌÁ÷»¯´²²ÉÓÃÅ·ÀÄ£ÐͲ¢ÐмÆËã³öÏÖÎÊÌâ
ÒѾÓÐ12È˻ظ´
¡¾ÇóÖú¡¿fluentÄ£ÄâÁ½¶ÎÁ÷»¯´²²ÉÓÃÅ·ÀºÍDPMÄ£ÐÍÎÊÌâ
ÒѾÓÐ11È˻ظ´
¡¾ÇóÖú¡¿¹ØÓÚÅ·À£À¸ñÀÊÈÕ·½³Ì£¨Euler-Lagrange equation£©¡¾Òѽâ¾ö¡¿
ÒѾÓÐ11È˻ظ´
¡¾ÇóÖú¡¿ÓÐËÖªµÀÕâ¸öÊÇʲô¸öÀëÉ¢·½Ê½°¡£¬¸µÀïÒ¶£¬Å·À....£¿
ÒѾÓÐ3È˻ظ´

dubo
½ð³æ (ÖøÃûдÊÖ)
- ³ÌÐòÇ¿Ìû: 4
- Ó¦Öú: 23 (СѧÉú)
- ¹ó±ö: 0.779
- ½ð±Ò: 569.2
- É¢½ð: 3220
- ºì»¨: 31
- ɳ·¢: 1
- Ìû×Ó: 1821
- ÔÚÏß: 349.5Сʱ
- ³æºÅ: 559371
- ×¢²á: 2008-05-17
- ÐÔ±ð: GG
- רҵ: ¸ß·Ö×Ó×é×°Ó볬·Ö×ӽṹ
¡ï ¡ï
ÓàÔó³É(½ð±Ò+2): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:32:39
ÓàÔó³É(½ð±Ò+2): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:32:39
|
²»¹ý£¬ÎÒ¶ÔÄãÓõÄÕâ¸öÌý¸ÐÐËȤ£¬ºÇºÇ #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); |
9Â¥2011-06-04 14:56:22
huycwork
½ð³æ (ÖøÃûдÊÖ)
- ³ÌÐòÇ¿Ìû: 22
- Ó¦Öú: 0 (Ó×¶ùÔ°)
- ½ð±Ò: 953
- É¢½ð: 663
- ºì»¨: 8
- ɳ·¢: 13
- Ìû×Ó: 1080
- ÔÚÏß: 264.1Сʱ
- ³æºÅ: 1257243
- ×¢²á: 2011-04-06
- רҵ: ½ðÈÚѧ
¡ï ¡ï ¡ï
dubo(½ð±Ò+1): ¶àл½»Á÷ 2011-06-04 14:42:08
ÓàÔó³É(½ð±Ò+2, ³ÌÐòÇ¿Ìû+1): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:31:23
dubo(½ð±Ò+1): ¶àл½»Á÷ 2011-06-04 14:42:08
ÓàÔó³É(½ð±Ò+2, ³ÌÐòÇ¿Ìû+1): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:31:23
|
C++´úÂ룺 #include 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 = i+i; j < BUFSZ; j+=i){ buf[j] += i; } } size_t d, s = 0; for(size_t i = 2; i < BUFSZ; ++i){ d = buf[i]; if(i == buf[d]){ if(i != d) s += i; } } return s; } int main(){ std::cout< |

2Â¥2011-06-02 22:34:29
wangww2011
ľ³æ (ÖøÃûдÊÖ)
- ³ÌÐòÇ¿Ìû: 13
- Ó¦Öú: 11 (СѧÉú)
- ½ð±Ò: 4023.1
- É¢½ð: 2709
- ºì»¨: 18
- ɳ·¢: 1
- Ìû×Ó: 1915
- ÔÚÏß: 1537.1Сʱ
- ³æºÅ: 772953
- ×¢²á: 2009-05-17
- ÐÔ±ð: GG
- רҵ: Äý¾Û̬ÎïÐÔ II £ºµç×ӽṹ
¡ï ¡ï ¡ï ¡ï
Сľ³æ(½ð±Ò+0.5):¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
dubo(½ð±Ò+1): ¶àл½»Á÷ 2011-06-04 14:43:26
ÓàÔó³É(½ð±Ò+2): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:31:33
Сľ³æ(½ð±Ò+0.5):¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
dubo(½ð±Ò+1): ¶àл½»Á÷ 2011-06-04 14:43:26
ÓàÔó³É(½ð±Ò+2): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:31:33
|
½á¹û 31626 elapsed time=0.020000 seconds. c´úÂë #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(sqrtn*sqrtn==n)sum-=sqrtn; return sum; } int euler21(int n){ int i,sum=0,tmp; for(i=3;i if(tmp!=i&&tmp } } return sum; } int main(void){ int i; TIMERSTART; printf("%d\n",euler21(10000)); TIMERSTOP; return 0; } |
3Â¥2011-06-03 13:22:19
libralibra
ÖÁ×ðľ³æ (ÖøÃûдÊÖ)
æôÆï½«¾ü
- ³ÌÐòÇ¿Ìû: 40
- Ó¦Öú: 817 (²©ºó)
- ½ð±Ò: 12914.1
- ºì»¨: 64
- Ìû×Ó: 2238
- ÔÚÏß: 287.3Сʱ
- ³æºÅ: 696514
- ×¢²á: 2009-02-05
- רҵ: ¼ÆËã»úÈí¼þ
¡ï ¡ï ¡ï ¡ï
Сľ³æ(½ð±Ò+0.5):¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
dubo(½ð±Ò+1): ¶àл½»Á÷ 2011-06-04 14:43:53
ÓàÔó³É(½ð±Ò+2, ³ÌÐòÇ¿Ìû+1): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:31:45
Сľ³æ(½ð±Ò+0.5):¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
dubo(½ð±Ò+1): ¶àл½»Á÷ 2011-06-04 14:43:53
ÓàÔó³É(½ð±Ò+2, ³ÌÐòÇ¿Ìû+1): ¹ÄÀø½»Á÷£¡ 2011-06-04 19:31:45
|
¿´±¿×¾µÄmatlab°É function result = euler21() tic; result = []; for i=1:10000 if d(i)~=i && d(d(i))==i result = [result,i]; end end result = sum(unique(result)); toc; end %% Let d(n) denote the sum of proper divisors of n (numbers less than n which divide evenly into n). % For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. % sub function to compute d(n), called by 21 and 23 function s = d(n) s = 0; for i=1:n-1 if mod(n,i)==0 s = s+i; end end end |

4Â¥2011-06-03 16:36:21














»Ø¸´´ËÂ¥
10