题目传送门 /* 题意:给出一个数,问是否有ai + bj + ck == x 二分查找:首先计算sum[l] = a[i] + b[j],对于q,枚举ck,查找是否有sum + ck == x */ #include <cstdio> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll a[MA…

Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 17036    Accepted Submission(s): 4337 Problem Description Give you three sequences of numbers A, B, C, then we give you a number…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2141 题目大意:查找是否又满足条件的x值. 这里简单介绍一个小算法,二分查找. /* x^2+6*x-7==y 输入y 求x 精确度为10^-5 0=<x<=10000 */ #include <iostream> #include <cstdio> using namespace std; int main (void) { double y; while(cin>…

Problem Description Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.   Input There are many cases. Every data cas…

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. Input There are many cases. Every data case is described as foll…

Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 9180    Accepted Submission(s): 2401 Problem Description Give you three sequences of numbers A, B, C, then we give you a number…

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. Input There are many cases. Every data case is described as foll…

题目传送门 题意:计算从1开始到第n个非完全平方数的开方和 分析:设第n个非完全平方数的值为a,x * x < a < (x+1) * (x+1),而且易得(tmp = sqrt (a) ) == x,a之前的非完全平方数的个数为a - tmp,所以可以二分查找a - tmp == n的a,然后模拟一下能计算出前a个数的开方和 收获:二分查找是个好方法 代码: /************************************************ * Author :Running…

题目传送门 题意:给出一些花开花落的时间,问某个时间花开的有几朵 分析:这题有好几种做法,正解应该是离散化坐标后用线段树成端更新和单点询问.还有排序后二分查找询问点之前总花开数和总花凋谢数,作差是当前花开的数量,放张图易理解: 还有一种做法用尺取法的思想,对暴力方法优化,对询问点排序后再扫描一遍,花开+1,花谢-1.详细看代码. 收获:一题收获很多:1. 降低复杂度可以用二分 2. 线段计数问题可以在端点标记1和-1 3. 离散化+线段树 终于会了:) (听说数据很水?) 代码1:离散化+线段树…

第一种:顺序查找法 中心思想:和数组中的值逐个比对! /* * 参数说明: * array:传入数组 * findVal:传入需要查找的数 */ function Orderseach(array,findVal){ var temp = false; //控制开关 for(var i =0;i<array.length;i++){ if(array[i] == findVal){ //逐个匹配是否相等 temp = true; //如果找到,temp设置为true; return i; //返…