count 0 for k 1 k n k* 2

fori=0ton docount[keyof(a[i])inpass j] fork=1to10docount[k

s=count_pi(n) s=0; for n=1:n s=s 1/n^2; end pi=s

voidmain(string[]args){intcount=0;intsum=0;intn=0;for(in

故a*不等于0,又根据上述公式aa*=0而a的秩小于n-1可知a的任意n-1阶余

100; cout a3; return0; } intmain(){ int i,j,k; int count

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