publicclassA{publicdoublefunc1(inti,intj){returni+j;}publicdoublefunc1(doublei,doublej){returni+j+10;}publicdoublefunc1(floati,floatj){returni+j+20;}publicdoublefunc1(longi,longj){returni+j+30;}}publicclassB௅୔ৠቹቹ뗐敔敔ৠቹ�敘ৠቹﲘǛ瘓轸ி㰰ǚ㰰ǚብ盂ீЀ￾￿￾￿痰௅ብ盂
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䕄Ȓ୰ቹء盜 ᤀ澈ঝ

下面程序的功能是用公式π2/6≈1/12+1/22+1/32+...+1/n2求π的近似值,直到最后一项的值小于10-6为止,请分析程序填空。#include#includemain(){longi=1;【1】pi=0;while(i*i>=1e6){pi=【2】;i++;}pi=sqrt(6.0*pi);printf("pi=%10.6f",pi);}