训练赛上一题，当时没做出来，Orz太弱了
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 2
28 ) 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).
题意：给出4个数列，从每个数列中选取一个元素，有多少种组合使得和为0？
思路：首先如果直接for暴力O(n^4)肯定时不行的，然后我们能想到如果让 c[i]和d[i]合并 a[i]与b[i]合并 找他们的相反数有多少个就可以了，
这里便出现了二分的应用：求 一个数 在一段序列中出现的次数。
我们可以先用 lower_bound 查找第一个大于x的数
函数lower_bound()在first和last中进行二分查找，如一个数组number序列1,2,2,4.lower_bound(2)后，返回的位置是1（下标）也就是2所在的位置,（只在单调递增的数组中适用）
函数upper_bound()返回的在前闭后开区间查找的关键字的上界，如一个数组number序列1,2,2,4.upper_bound(2)后，返回的位置是3（下标）也就是4所在的位置,
代码：
#include#include #include <string.h> #include #include <string> #include #include #include #include #include #include #include