Description
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The SUM problem can be formulated as follows: given four lists A, B, C, D<tex2html_verbatim_mark> of integer values, compute how many quadruplet (a, b, c, d ) AxBxCxD<tex2html_verbatim_mark> are such that a + b + c + d = 0<tex2html_verbatim_mark> . In the following, we assume that all lists have the same size n<tex2html_verbatim_mark> .
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The first line of the input file contains the size of the lists n<tex2html_verbatim_mark> (this value can be as large as 4000). We then have n<tex2html_verbatim_mark> lines containing four integer values (with absolute value as large as 228<tex2html_verbatim_mark> ) that belong respectively to A, B, C<tex2html_verbatim_mark> and D<tex2html_verbatim_mark> .
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
For each input file, your program has to write the number quadruplets whose sum is zero.
Sample Input
1 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
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).
如果按照原始方法,进行四重循环,毫无疑问是会超时的,只能使用方法把四重循环变成两重,可是每一个数最大可达2的28次方,使用short开数组也是行不通的,做一个大整数的hash。。不会,只能使用数组保存值,之后再进行查找了。。。
#include"iostream"
#include"cstring"
#include"algorithm"
#include"map"
#include"cmath"
using namespace std;
const int maxn=4000+10;
int book[16000010];
int a[maxn],b[maxn],c[maxn],d[maxn];
int main()
{
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
memset(book,0,sizeof(book));
for(int i=0;i<n;i++)
cin>>a[i]>>b[i]>>c[i]>>d[i];
int f=0;
for(int ia=0;ia<n;ia++)
for(int ib=0;ib<n;ib++)
{
book[f++]=-(a[ia]+b[ib]);
}
int sum=0;
sort(book,book+f);
for(int ic=0;ic<n;ic++)
for(int id=0;id<n;id++)
{
int temp=c[ic]+d[id];
int x1=lower_bound(book,book+f,temp)-book;
int x2=upper_bound(book,book+f,temp)-book;
sum+=x2-x1;
}
cout<<sum<<endl;
if(T) cout<<endl;
}
return 0;
}