<span style="color:#3333ff;">/*A - 二分法 基础
Time Limit:15000MS Memory Limit:228000KB 64bit IO Format:%I64d & %I64u
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Status
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 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).
By Grant Yuan
2014.7.14
二分
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
long long a[4002],b[4002],c[4002],d[4002];
int n;
long long num1[16000004];
long long num2[16000004];
int top;
long long sum;
int binary(long k,int left,int right)
{
int i;
while(left<=right){
int mid=(left+right)/2;
int num=0;
if(num2[mid]==k)
{
num=1;
for(i=mid-1;i>=0&&num2[i]==k;i--) num++;
for(i=mid+1;i<n*n&&num2[i]==k;i++) num++;
return num;
}
else if(num2[mid]>k)
right=mid-1;
else left=mid+1;
}
return 0;
}
int main()
{
cin>>n;
long long flag;
for(int i=0;i<n;i++)
cin>>a[i]>>b[i]>>c[i]>>d[i];
top=-1;
sum=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
num1[++top]=a[i]+b[j];
num2[top]=c[i]+d[j];
}
sort(num1,num1+top+1);
sort(num2,num2+top+1);
for(int i=0;i<=top;i++)
{
flag=-num1[i];
sum+=binary(flag,0,top+1);
}
cout<<sum<<endl;
return 0;
}
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