POJ 2785 4 Values whose Sum is 0

Posted 2020-09-30 FTA_Macro

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 22691		Accepted: 6869
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

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).

Source

Southwestern Europe 2005

 

题目大意：给出四列数A,B,C,D，每列中选出一个数啊a,b,c,d，问有多少种情况可以使得a+b+c+d=0。

 

大致思路：直接用四个循环超时是肯定的，这时可以用折半的方法，先求a[i]+b[i]，再求c[i]+d[i]，用二分找出c[i]+d[i]中满足的情况。详见代码（本题还能用hash做）

 

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 4005
int a[maxn],b[maxn],c[maxn],d[maxn];
int ab[maxn*maxn],cd[maxn*maxn];
int main()
{
    int i,j,N,lo,hi,mid,cou=0,p=0;
    cin>>N;
    for(i=0;i<N;i++)
        cin>>a[i]>>b[i]>>c[i]>>d[i];
    for(i=0;i<N;i++)
        for(j=0;j<N;j++)
        {
            ab[p]=a[i]+b[j];
            p++;
        }
        sort(ab,ab+N*N);
    p=0;
    for(i=0;i<N;i++)
        for(j=0;j<N;j++)
        {
            cd[p]=c[i]+d[j];
            p++;
        }
        sort(cd,cd+N*N);
    for(i=0;i<p;i++)
    {
        lo=0;hi=p;//从p-1和p开始二分都行
        while(lo<hi)//这个循环会找到一个与要找的数相等的最小下标
        {
            mid=(lo+hi)/2;
            if(ab[mid]<-cd[i])//因为是排好序的，所以可以直接利用数组下标二分
                lo=mid+1;
            else hi=mid;
        }
        while(ab[lo]==-cd[i]&&lo<p)//判断lo和lo之后是否相等，因为可能会有数据相同，注意lo<p！！！
        {
            cou++;
            lo++;
        }
    }
    cout<<cou<<endl;
    return 0;
}