原题:http://www.spoj.com/problems/FACVSPOW/ 题目大意: 给正整数a, 求最小的正整数n, 使得n!>a^n 分析: 我看下面留言都说不计算fact, 用数学方法. 我就想两边取log, 左边的阶乘就是log(1)+log(2)….log(n), 右边的power就是n*log(a), 然后我写了下面的code public void solve(int testNumber, InputReader in, OutputWriter out) { int n = in.readInt(); double power = 0; double fact = 0; int cur = 2; double temp = Math.log(n); while (true) { fact += Math.log(cur); // power = temp * cur; if (fact […]