POJ - 2785

4 Values whose Sum is 0

Time Limit: 15000MS		Memory Limit: 228000K
Total Submissions: 29083		Accepted: 8820
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).

 

两种方法：

方法一：利用函数的增减性

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <map>
using namespace std;
int a[4001],b[4001],c[4001],d[4001];
int ab[4000*4000+1],cd[4000*4000+1];
int main(){
    int n;
    while(~scanf("%d",&n)){
        int sum=0,num,x=n*n-1;
        for(int i=0;i<n;i++)
            scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
        int l=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++)
                ab[l++]=a[i]+b[j];
        }
        l=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++)
                cd[l++]=c[i]+d[j];
        }
        sort(ab,ab+n*n);
        sort(cd,cd+n*n);

        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){
                sum++;
                num--;

            }

        }
        printf("%d\n",sum);
    }

    return 0;
}

方法二：二分查找（慢）

#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;
}