val ini sum following ont hose tdi size pac

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). 題目意思就是在一個4*N的矩陣裏面 不同的行找出相加為0的數量； 暴力搜索的話四重循環肯定是不行的，但是如果把他分成兩部分來搜，復雜度就從n^4到了2*n^2 數據最大4000 所以完全不會超時

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int a[4005],b[4005],c[4005],d[4005];
int ab[16000005],cd[16000005];
int main()
{
    int n,k=0,l=0,count=0;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i]>>b[i]>>c[i]>>d[i];
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            ab[k++]=a[i]+b[j];
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            cd[l++]=c[i]+d[j];
    sort(ab,ab+k);
    sort(cd,cd+l);
    int x=n*n-1,num=0;
    for(int i=0;i<n*n;i++){
        while(x>=0&&ab[i]+cd[x]>0)
            x--;
        if(x<0)
            break;
        num=x;
        while(ab[i]+cd[num]==0&&num>=0){
            count++;
            num--;
        }
    }
    cout<<count<<endl;
    return 0;
}

二分

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[4001],b[4001],c[4001],d[4001];
int ab[4000*4000+1],cd[4000*4000+1];
int k2;
int check(int x)
{
    int left=1,right=k2-1,mid;
    while (left<=right)
    {
        mid=(left+right)/2;
        if (x==cd[mid])
        {
            int w=0,e=mid;
            while (x==cd[e]&&e<k2)
                e++,w++;
            e=mid-1;
            while (x==cd[e]&&e>0)
                e--,w++;
            return w;
        }
        else if (x<cd[mid])
            right=mid-1;
        else
            left=mid+1;
    }
    return 0;
}
int main()
{
    int t,i,j,q;
    while (~scanf("%d",&t))
    {
        for (i=1;i<=t;i++)
            scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
        memset(ab,0,sizeof(ab));
        memset(cd,0,sizeof(cd));
        int k1=1,sum=0;
        k2=1;
        for (i=1;i<=t;i++)
        {
            for (j=1;j<=t;j++)
            {
                ab[k1++]=a[i]+b[j];
                cd[k2++]=-(c[i]+d[j]);
            }
        }
        sort(cd+1,cd+k2);
        for (i=1;i<k1;i++)
            sum+=check(ab[i]);
        printf("%d\n",sum);
    }
    return 0;
}

　　

POJ 2785 (暴力搜索&二分)