1
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>> SPIKA 1: Hebu kuandika mpango huo
humshauri mtumiaji kwa chanya

2
00:00:02,830 --> 00:00:05,950
integer, n, na kisha Prints nje
Jumla ya namba zote

3
00:00:05,950 --> 00:00:07,980
kati ya 1 na n.

4
00:00:07,980 --> 00:00:10,580
Naam, hapa tuna kuu, ambayo nimekuwa
tayari imeandikwa mapema.

5
00:00:10,580 --> 00:00:13,520
Na taarifa hapa juu ya
kuu, mimi kutangaza int n.

6
00:00:13,520 --> 00:00:16,079
>> Mimi basi, ndani ya kufanya wakati
kitanzi, kwanza magazeti nje

7
00:00:16,079 --> 00:00:17,530
integer chanya, tafadhali.

8
00:00:17,530 --> 00:00:21,070
Kisha mimi kuendelea kupata integer kutoka
user kwa kupata maktaba CS50

9
00:00:21,070 --> 00:00:22,070
int kazi.

10
00:00:22,070 --> 00:00:26,410
Na kisha katika wakati hali yangu hapa, mimi
kuhakikisha kwamba n ni kubwa zaidi kuliko au

11
00:00:26,410 --> 00:00:30,480
sawa na 1 kabla ya mimi kwa kweli kuendelea
kufanya kitu kwa thamani hiyo.

12
00:00:30,480 --> 00:00:31,520
>> Je, mimi kufanya ijayo?

13
00:00:31,520 --> 00:00:34,690
Naam, mimi wito kazi kwamba mimi nina kwenda
kuwaita sigma, mwakilishi wa

14
00:00:34,690 --> 00:00:37,700
mji mkuu wa sigma kwamba unaweza kuwa
alikumbuka kutoka madarasa math kwamba

15
00:00:37,700 --> 00:00:40,860
inaonyesha kwamba unataka kwa jumla kitu
kutoka thamani mmoja hadi mwingine.

16
00:00:40,860 --> 00:00:44,540
Na chochote kwamba kazi ya anarudi kama
thamani yake kurudi, mimi nina kwenda kuhifadhi

17
00:00:44,540 --> 00:00:46,500
katika variable kuitwa jibu.

18
00:00:46,500 --> 00:00:50,280
>> Hatimaye, katika line yangu ya mwisho katika kuu, mimi nina
kwenda magazeti nini jibu ni.

19
00:00:50,280 --> 00:00:52,840
Bila shaka, sisi bado kutekelezwa
hii sigma kazi.

20
00:00:52,840 --> 00:00:54,590
Hivyo ni jinsi gani sisi kwenda juu ya kufanya hivyo?

21
00:00:54,590 --> 00:00:58,040
>> Naam, chini ya file yangu, mimi nina
kwenda kuendelea na kutangaza kazi

22
00:00:58,040 --> 00:00:59,450
kwamba anarudi int.

23
00:00:59,450 --> 00:01:01,630
Na mimi nina kwenda kuwaita
kwamba kazi ya sigma.

24
00:01:01,630 --> 00:01:06,340
Na mimi nina kwenda bayana kwamba kama pembejeo
kazi ambayo anapokea pia int.

25
00:01:06,340 --> 00:01:09,800
Na mimi itabidi kuiita tu, kwa kuwa
tofauti, m badala ya n.

26
00:01:09,800 --> 00:01:12,120
Lakini tungeweza kuitwa kuwa ni
wengi chochote tunatarajia.

27
00:01:12,120 --> 00:01:14,930
>> Ndani ya mwili wa hii kazi mimi nina
kwenda kuendelea kutumia ukoo

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00:01:14,930 --> 00:01:16,420
kujenga, yaani kitanzi.

29
00:01:16,420 --> 00:01:19,010
Lakini mimi nina pia kwenda kufanya kidogo ya sanity
kuangalia ili kuhakikisha kuwa

30
00:01:19,010 --> 00:01:22,340
user haitoi mimi na idadi
kwamba mimi si kutarajia.

31
00:01:22,340 --> 00:01:28,010
Hasa, mimi nina kwenda kufanya kama m ni
chini ya 1 na, kwa kiasi fulani kiholela,

32
00:01:28,010 --> 00:01:31,280
Mimi tu kwenda na kurudi 0 kama
idadi ni si mazuri

33
00:01:31,280 --> 00:01:32,800
integer kama mimi kutarajia.

34
00:01:32,800 --> 00:01:36,920
>> Basi mimi nina kwenda kutangaza variable
aitwaye Jumla na initialize 0.

35
00:01:36,920 --> 00:01:40,810
Hii hatimaye kuhifadhi Jumla ya
wote wa idadi ya kati ya 1 na m.

36
00:01:40,810 --> 00:01:43,550
Na kisha mimi naenda kutumia ukoo
mbele kitanzi kujenga.

37
00:01:43,550 --> 00:01:50,272
Kwa int i anapata 1, i ni chini ya
au sawa na m, i pamoja na plus.

38
00:01:50,272 --> 00:01:54,010
Na kisha, ndani ya mwili wa hii
kitanzi, mimi nina tu kwenda kufanya Jumla

39
00:01:54,010 --> 00:01:56,350
sawa na Jumla pamoja na i.

40
00:01:56,350 --> 00:02:01,900
Au, zaidi tu, jumla pamoja na sawa na i,
ambayo inafikia matokeo hayo.

41
00:02:01,900 --> 00:02:04,810
>> Na kisha mwisho, mimi haja ya kurudi
kiasi kwamba nimekuwa computed.

42
00:02:04,810 --> 00:02:07,640
Basi, mimi kuongeza katika kurudi jumla.

43
00:02:07,640 --> 00:02:08,560
>> Sasa mimi nina si kufanyika bado.

44
00:02:08,560 --> 00:02:11,360
Mimi haja ya kufundisha C kwamba hii
kazi kweli lipo.

45
00:02:11,360 --> 00:02:14,400
Na hivyo atop file yangu mimi nina kwenda kutangaza
nini tumekuwa aitwaye kazi

46
00:02:14,400 --> 00:02:18,270
mfano, sawa na sahihi
kwamba mimi kutumika wakati kufafanua kazi

47
00:02:18,270 --> 00:02:19,250
wakati iliyopita.

48
00:02:19,250 --> 00:02:22,450
>> Hasa, tu juu ya kuu,
Mimi nina kwenda aina int

49
00:02:22,450 --> 00:02:26,080
sigma, int m, semicolon.

50
00:02:26,080 --> 00:02:29,240
Si kutekeleza kazi
tena, kuutangaza tu.

51
00:02:29,240 --> 00:02:32,800
Kama mimi sasa kuokoa, kukusanya, na kukimbia hii
mpango, hebu angalia nini mimi kupata.

52
00:02:32,800 --> 00:02:37,460
Kufanya sigma 0 dot kufyeka sigma 0.

53
00:02:37,460 --> 00:02:41,050
Na sasa hebu kutoa integer chanya
kama 2, ambayo anipe

54
00:02:41,050 --> 00:02:45,920
tatu, kwa sababu maadili kati ya
1 na 2 ni 1 plus 2 ni sawa na 3.

55
00:02:45,920 --> 00:02:47,300
Na kwa kweli, kwamba ni nini mimi kupata.

56
00:02:47,300 --> 00:02:49,940
>> Hebu kukimbia tena, hii
wakati na, kusema, 3.

57
00:02:49,940 --> 00:02:53,470
Hivyo ni lazima kupata 1 plus 2 pamoja na
3 anipe 6.

58
00:02:53,470 --> 00:02:54,740
Na hakika, mimi kupata 6.

59
00:02:54,740 --> 00:02:57,380
>> Na hebu jaribu thamani moja mwisho, wanasema 50.

60
00:02:57,380 --> 00:03:01,160
Na 1275 ni jibu letu.

61
00:03:01,160 --> 00:03:02,253