這個題目同上道二分的題目一樣，只是把數字分成兩堆，在排序用二分的思想，但是必須明白，不是在找到一個的條件的情況下跳出，可能只有重復的。這裡是重點。繼續我的二分之路，我的時間復雜度很高，效率不怎麼好。看了下別人的代碼，好像都是用hash，下一站就是hash了。
4 Values whose Sum is 0
Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 11586 Accepted: 3249
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
代碼：
[cpp]
<span style="font-family:FangSong_GB2312;font-size:18px;">#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
#define maxn 4004
using namespace std;
int map1[maxn*maxn];
int map2[maxn*maxn];
int a[maxn],b[maxn],c[maxn],d[maxn];
int main()
{
int n,i,j,k,sum,p;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
map1[i*n+j]=a[i]+b[j];
for(i=0;i<n;i++)
for(j=0;j<n;j++)
map2[i*n+j]=c[i]+d[j];
sort(map1,map1+n*n);
sort(map2,map2+n*n);
sum=0;
p=n*n-1;
for(i=0;i<n*n;i++)
{
while(p>=0&&map1[i]+map2[p]>0) p--;
if(p<0) break;
int temp=p;
while(temp>=0&&map1[i]+map2[temp]==0)
{
sum++; temp--;
}
}
printf("%d\n",sum);
//system("pause");
return 0;
}
</span>
