4 Values whose Sum is 0
Time Limit: 15000MS   Memory Limit: 228000K
Total Submissions: 25615   Accepted: 7697
Case Time Limit: 5000MS

Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 228 ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

Source


题目大意:给4列数,每列有n个数字,从每列中取一个数,求四个数相加为零的情况有多少种。
思路:用两个数组,分别记录左边两列和右边两列的数相加的和,再将其中一个数组从小到大排列,将另一数组遍历一边,用二分的方法判断个数。
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[4005][4], sum1[16000001], sum2[16000001];
int main()
{
	int n, mid;

	while (~scanf("%d", &n))
	{
		int k = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d %d %d %d", &a[i][0], &a[i][1], &a[i][2], &a[i][3]);
		}
		for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
		{
			sum1[k] = a[i][0] + a[j][1];
			sum2[k++] = a[i][2] + a[j][3];
		}
		sort(sum1, sum1 + k);

		int num = 0;
		for (int i = 0; i < k; i++)
		{
			int left = 0, right = k - 1;
			while (left <= right)
			{
				mid = (left + right) / 2;
				if (sum2[i] + sum1[mid] == 0)
				{
					num++;
					for (int j = mid + 1; j < k; j++)
					{
						if (sum2[i] + sum1[j] == 0)
							num++;
						else
							break;
					}
					for (int j = mid - 1; j >= 0; j--)
					{
						if (sum2[i] + sum1[j] == 0)
							num++;
						else
							break;
					}
					break;
				}
				if (sum2[i] + sum1[mid] > 0)
					right = mid - 1;
				else
					left = mid + 1;
			}
		}
		printf("%d\n", num);
	}
	return 0;
}



POJ - 2785 4 Values whose Sum is 0 二分的更多相关文章

  1. POJ - 2785 - 4 Values whose Sum is 0 - 二分折半查找

    2017-08-01 21:29:14 writer:pprp 参考:http://blog.csdn.net/piaocoder/article/details/45584763 算法分析:直接暴力 ...

  2. POJ 2785 4 Values whose Sum is 0(想法题)

    传送门 4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 20334   A ...

  3. POJ 2785 4 Values whose Sum is 0

    4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 13069   Accep ...

  4. POJ 2785 4 Values whose Sum is 0(折半枚举+二分)

    4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 25675   Accep ...

  5. POJ 2785 4 Values whose Sum is 0(暴力枚举的优化策略)

    题目链接: https://cn.vjudge.net/problem/POJ-2785 The SUM problem can be formulated as follows: given fou ...

  6. POJ 2785 4 Values whose Sum is 0(哈希表)

    [题目链接] http://poj.org/problem?id=2785 [题目大意] 给出四个数组,从每个数组中选出一个数,使得四个数相加为0,求方案数 [题解] 将a+b存入哈希表,反查-c-d ...

  7. POJ 2785 4 Values whose Sum is 0 Hash!

    http://poj.org/problem?id=2785 题目大意: 给你四个数组a,b,c,d求满足a+b+c+d=0的个数 其中a,b,c,d可能高达2^28 思路: 嗯,没错,和上次的 HD ...

  8. poj 2785 4 Values whose Sum is 0(折半枚举(双向搜索))

    Description The SUM problem can be formulated . In the following, we assume that all lists have the ...

  9. [POJ] 2785 4 Values whose Sum is 0(双向搜索)

    题目地址:http://poj.org/problem?id=2785 #include<cstdio> #include<iostream> #include<stri ...

随机推荐

  1. a版本冲刺第五天

    队名:Aruba   队员: 黄辉昌 李陈辉 林炳锋 鄢继仁 张秀锋 章  鼎 运动会这几天两位同学准备比赛也确实比较忙,两位同学又刚好有事回家去了,由于之前git解决一次冲突后,远程我们还不能很好地 ...

  2. Java 异常java.lang.IllegalArgumentException: Illegal group reference

    当字符串方法replaceAll()中替换字符含有特殊字符$如, String test = "<StreamingNo>abc</StreamingNo>" ...

  3. ToJson

    /// <summary> /// DataTable转成Json /// </summary> /// <param name="jsonName" ...

  4. 【leetcode】Remove Duplicates from Sorted List II (middle)

    Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numb ...

  5. 【转】【DP_树形DP专辑】【9月9最新更新】【from zeroclock&#39;s blog】

    树,一种十分优美的数据结构,因为它本身就具有的递归性,所以它和子树见能相互传递很多信息,还因为它作为被限制的图在上面可进行的操作更多,所以各种用于不同地方的树都出现了,二叉树.三叉树.静态搜索树.AV ...

  6. asp.net mvc JQGrid

    http://mikedormitorio.azurewebsites.net/BlogPost/jqgrid-series-part-1-loading-data-to-a-jqgrid-on-an ...

  7. NotePad++更改背景颜色

    白色的编辑框看得眼睛不舒服,怎么样更改NotePad++的背景颜色使眼睛更舒服些? 1.设置--语言格式设置 2.设置背景色 “背景色”一栏,选择背景色颜色   “使用全局背景色”一栏要打上√,否则无 ...

  8. C++Primer第5版学习笔记(四)

    C++Primer第5版学习笔记(四) 第六章的重难点内容         你可以点击这里回顾第四/五章的内容       第六章是和函数有关的知识,函数就是命名了的代码块,可以处理不同的情况,本章内 ...

  9. 一步步教你如何源码编译Recovery

    *1 准备Ubuntu作为您的操作系统,笔者的版本是12.04_amd64. *2 准备 Android 源码的编译环境,主要是安装一些编译用到的lib库,以及同步源码的一些工具 ,如GIT,CURL ...

  10. day3、Linux快捷键及vim命令快捷键

    Linux命令行快捷键 快捷键: tab键  自动补全路径 目录  名字,  自动不全命令 快捷键: ctrl +l(小写) 清屏 . ctrl +c 取消当前操作 快捷键: ctrl +d(小写) ...