1 00:00:00,000 --> 00:00:03,110 >> SPIKA 1: Katika toleo la mwisho la sigma, mimi kutekelezwa kile ambacho linaweza kujulikana 2 00:00:03,110 --> 00:00:06,570 ufumbuzi iterative, ambapo Nilikuwa mbele kitanzi kuhesabu up yote ya 3 00:00:06,570 --> 00:00:09,720 idadi kati ya 1 na m, baada ya hapo kurudi jumla. 4 00:00:09,720 --> 00:00:12,560 >> Lakini zinageuka tunaweza kutumia mwingine mbinu ya kutekeleza hiyo 5 00:00:12,560 --> 00:00:15,120 kazi, mbinu inayojulikana kama recursion. 6 00:00:15,120 --> 00:00:19,360 kazi ya kujirudia, hivyo kusema, ni moja tu kwamba wito yenyewe. 7 00:00:19,360 --> 00:00:21,290 Sasa, katika na wenyewe, kwamba inaweza kuwa tatizo. 8 00:00:21,290 --> 00:00:24,500 Kama kazi tu wito yenyewe ambayo wito yenyewe ambayo wito yenyewe, 9 00:00:24,500 --> 00:00:26,080 mchakato kwamba wanaweza BoT milele mwisho. 10 00:00:26,080 --> 00:00:30,490 Lakini kwa muda mrefu kama sisi ni pamoja na kinachojulikana msingi kesi, hali ambayo kuhakikisha 11 00:00:30,490 --> 00:00:34,930 kwamba katika baadhi ya hali hatuwezi kuwaita wenyewe, kuwa mchakato wa vinginevyo 12 00:00:34,930 --> 00:00:37,070 looping usio lazima zitakoma. 13 00:00:37,070 --> 00:00:39,180 >> Hebu sasa reimplement sigma kama ifuatavyo. 14 00:00:39,180 --> 00:00:43,810 Kama n ni chini ya au sawa na 0, mimi nina tu, na kwa kiasi fulani kiholela, 15 00:00:43,810 --> 00:00:45,670 kwenda na kurudi 0. 16 00:00:45,670 --> 00:00:49,370 Mwingine nini mimi kwenda kufanya ni kweli compute sigma kwa chanya int 17 00:00:49,370 --> 00:00:50,460 kwamba nimekuwa mitupu. 18 00:00:50,460 --> 00:00:52,050 >> Sasa, ni nini sigma ya m? 19 00:00:52,050 --> 00:00:55,480 Naam, sigma ya m ni, bila shaka, Jumla ya 1 ya juu kupitia m. 20 00:00:55,480 --> 00:00:58,820 Lakini kama sisi kufikiri juu yake kwa njia nyingine, ni tu Jumla ya m pamoja na m 21 00:00:58,820 --> 00:01:02,560 minus 1 pamoja na m bala 2 na kadhalika, njia yote chini ya 1. 22 00:01:02,560 --> 00:01:08,080 Hivyo kwa mantiki kwamba, inaonekana kwamba Mimi naweza tu kurudi m plus. 23 00:01:08,080 --> 00:01:10,210 >> Na kisha mimi haja m minus 1 pamoja na m bala 2. 24 00:01:10,210 --> 00:01:13,470 Lakini nina kazi ambayo inaweza kutoa mimi hasa kwamba jibu, yaani 25 00:01:13,470 --> 00:01:16,340 sigma ya m minus 1. 26 00:01:16,340 --> 00:01:19,670 >> Sasa, wito mwenyewe kwa njia hii haina kuonekana kama wazo bora. 27 00:01:19,670 --> 00:01:22,610 Kwa sababu kama sigma wito sigma ambayo inatoa wito sigma ambayo inatoa wito sigma, wewe 28 00:01:22,610 --> 00:01:24,480 kudhani kwamba mchakato huu wanaweza milele mwisho. 29 00:01:24,480 --> 00:01:27,720 Lakini kwamba ni kwa nini tulikuwa kinachojulikana msingi kesi juu ya kazi hii. 30 00:01:27,720 --> 00:01:31,540 kama hali ya kwamba hundi kama m ni chini ya au sawa na 0 Sitakuja 31 00:01:31,540 --> 00:01:32,610 kuwaita mwenyewe. 32 00:01:32,610 --> 00:01:37,010 Mimi badala kwenda na kurudi 0, ambayo kwa upande mwingine ni kwenda kuongezwa kwa 33 00:01:37,010 --> 00:01:39,950 idadi uliopita kwamba nimekuwa summing up, na hivyo kuacha hii 34 00:01:39,950 --> 00:01:41,740 mchakato vinginevyo kubwa. 35 00:01:41,740 --> 00:01:43,710 >> Hebu sasa kuona kama hii mpya utekelezaji wa kazi. 36 00:01:43,710 --> 00:01:46,510 Hebu kuokoa, kukusanya, na kuendesha mpango huu. 37 00:01:46,510 --> 00:01:50,640 Kufanya sigma 1 dot kufyeka sigma 1. 38 00:01:50,640 --> 00:01:52,900 Na hebu kutoa ni pamoja na idadi sawa mbele. 39 00:01:52,900 --> 00:01:55,520 2, ambayo inapaswa hopefully nipe 3. 40 00:01:55,520 --> 00:01:58,970 Hebu kutoa ni pamoja na 3, ambayo lazima hopefully nipe 6. 41 00:01:58,970 --> 00:02:03,480 Na hebu hatimaye kutoa ni pamoja na 50, ambayo kwa hakika anitiaye 1275. 42 00:02:03,480 --> 00:02:06,130