2^1000000000000000怎么计算结果位数
设2^N是k位数,则:10^(k+1)>2^N>10^k取对数得:k+1 > log(2^N)=Nlog2 > k所以:k=Nlog2整数位+1所以:k=[Nlog2]+1
N=100000000000000,log2=0.3010299956639812
k=[Nlog2]+1=int(301029995663981.2)+1=301029995663982