C/C++
摘要:下列程序输出:67 解析: 解法1: f(n)=0, cnt自加一次,f(n)=1, cnt自加一次,即: n=0 cnt = 1; n=1 cnt = 1; n=2 cnt = f(1) + f(0) = 1+1+1 = 3; n=3 cnt = f(2) + f(1) = 3+1+1 = 5;
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posted @ 2018-05-22 12:23
posted @ 2018-05-22 12:23