转载请注明出处:http://blog.csdn.net/u012860063 题目链接:id=3903">http://poj.org/problem?id=3903 Description The world financial crisis is quite a subject. Some people are more relaxed while others are quite anxious. John is one of them. He is very concerned
题意:给定 n 堆数,然后有 m 个话询问,问你在哪一堆里. 析:这个题是一个二分题,但是有一个函数,可以代替写二分,lower_bound. 代码如下: #include<bits/stdc++.h> using namespace std; typedef long long LL; const int maxn = 1e5 + 5; int a[maxn]; int main(){ int n, m; cin >> n; for(int i = 1; i <= n; +
1.lower_bound:查找序列中的第一个出现的值大于等于val的位置 这个序列中可能会有很多重复的元素,也可能所有的元素都相同,为了充分考虑这种边界条件,STL中的lower_bound算法总体上是才用了二分查找的方法,但是由于是查找序列中的第一个出现的值大于等于val的位置,所以算法要在二分查找的基础上做一些细微的改动. 首先是我修改数据结构课本上的二分查找实现的lower_bound算法: int lower_bound(vector<int> array, int size, in
Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14635 Accepted Submission(s): 4158 Problem Description JGShining's kingdom consists of 2n(n is no mo
Search Insert Position Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the array. Here are few examples. [1,3,5
本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie lower_bound(应用于有序区间) -------------------------------------------------------------------------------------------------------------------------- 描写叙述:二分查找,返回一个迭代器指向每个"不小于 value "的元素, 或 value 应