题干：

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).

解题报告：

   这题要求和为0，显然直接暴力是不存在的，但是还是有解题线索的，比如说一共就四个数，显然可以折半啊！！相同思路的遇到过一次了吧？题目名叫“暴力AC”，写过那篇博客但是具体哪场比赛忘记了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll a[4005],b[4005],c[4005],d[4005];
int tot1=0,tot2=0;
ll ans1[16000000],ans2[16000000];
int get(ll x) {return lower_bound(ans2+1,ans2+tot2+1,x) - ans2;
}
int main()
{int n;while(cin>>n) {for(int i = 1; i<=n; i++) {scanf("%lld%lld%lld%lld",a+i,b+i,c+i,d+i);}tot1=tot2=0;for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {ans1[++tot1] = a[i] + b[j];ans2[++tot2] = c[i] + d[j];}}ll ans = 0;//	sort(ans1+1,ans1+tot1+1);sort(ans2+1,ans2+tot2+1);for(int i = 1; i<=tot1; i++) {if(binary_search(ans2+1,ans2+tot2+1,-ans1[i]))ans += get(-ans1[i]+1) - get(-ans1[i]);}printf("%lld\n",ans);	}return 0 ;}

总结：

  本来想着可能会炸空间，是不是需要离散化一下，但是后来一想发现不可以这样，相反，如果离散化了这题就错了，因为他要计算的是次数，所以不能去重啊！！而且枚举的时候不能直接加binarysearch，而是应该先判断是否存在，如果存在的话加上所有存在的个数，可以按照我这样写，也可以一个lowerbound一个upperbound、。、、、

事实证明  longlong的1e7是开的下的，这次提交一共用了97796kB内存而已、、、。