原题: http://codeforces.com/contest/569/problem/A 题目大意: 下载一个T秒音乐, 当S秒的时候开始播放, 如果播放到没有下载的地方, 就重新播放, 下载速度是q/q-1, 问播放几次能下载完毕. 分析: 呵呵, 居然不会做, 我看了半天没明白为什么是q-1没有用到. 比赛的时候还有个解释, 说下载速度是q/q-1, 即 q是现实时间, q-1就是下载track时间. public void solve(int testNumber, InputReader in, OutputWriter out) { int t = in.readInt(); int s = in.readInt(); int q = in.readInt(); int n = 0; while ( s < t) { s *= q; n++; } out.print(n); }
private static Solution s = null; public static Solution getInstance() { // write your code here if(s == null){ s = new Solution(); return s; } else return s; }
public int fibonacci(int n) { // write your code here int first = 0; if(n == 1) return first; int second = 1; if(n == 2) return second; int third = 0; for(int i = 3; i <=n; i++) { third = first+ second; first = second; second = third; } return third; }
public ArrayList<ArrayList<Integer>> threeSum(int[] numbers) { // write your code here ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); if(numbers.length == 0 || numbers == null) return res; Arrays.sort(numbers); for(int i = 0 ; i < numbers.length – 2; i++) { if(i != 0 && numbers[i] == numbers[i-1]) continue; int l = i+1; int r = numbers.length – 1; […]
public int threeSumClosest(int[] numbers ,int target) { // write your code here Arrays.sort(numbers); if(numbers.length == 0 || numbers == null) return 0; int close = Integer.MAX_VALUE; for(int i = 0; i < numbers.length – 2; i++) { int j = i + 1; int k = numbers.length – 1; while(j < k) { int sum […]
public int firstMissingPositive(int[] A) { // write your code here if(A == null || A.length == 0) return 1; int i = 0; int n = A.length; while(i < n) { if(A[i] >= 0 && A[i] < n && A[A[i]] != A[i]) swap(A, i, A[i]); else i++; } int k = 1; while(k < n […]
public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) { ArrayList<Long> res = new ArrayList<Long>(); if(A.size()==0 || A == null) return res; long p = 1; for(int i = 0 ; i < A.size(); i++) { res.add(p); p *= A.get(i); } p = A.get(A.size()-1); for(int i = A.size()-2; i >= 0; i–) { res.set(i,p*res.get(i)); p *= A.get(i); } return res; […]