随笔分类 - 贪心+排序处理
摘要:题目链接:http://codeforces.com/contest/351/problem/A算法思路:2n个整数,一半向上取整,一半向下。我们设2n个整数的小数部分和为sum.ans = |A - B|;sum = A +(n-b)-B;所以ans = |sum - (n-b)|; 只有b未知,只需要枚举一下b就得到答案。#include#include#include#include#includeusing namespace std;const int maxn = 2005;const double eps = 1e-12;int dcmp(double x){ if(fab...
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