文章标题POJ 2785：4 Values whose Sum is 0？（二分）

4 Values whose Sum is 0

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).

题意：题目要我们从四个表中各找一个数相加的和为0。
分析：直接暴力破解肯定不行，可先对四个数表分成两组进行枚举相加，然后其中一个枚举，另一个二分，记住，记录重复的情况。
代码：

#include<iostream>#include<cstdio>#include<cstring>#include<math.h>#include<algorithm>using namespace std;int a[4005],b[4005],c[4005],d[4005];int sum1[4005*4005],sum2[4005*4005];int bs (int a[],int key,int k){    int lo=0,hi=k-1;    int mi,ans=hi+1;    while (lo<=hi){        mi=((hi-lo)>>1)+lo;        //if (a[mi]==key)return mi;        if (a[mi]<key)lo=mi+1;        else {            hi=mi-1;ans=min(ans,mi);            }    }    if (ans==hi+1)        return -1;    return ans;}int main (){    int n;    while (scanf ("%d",&n)!=EOF){        for (int i=0;i<n;i++){            scanf ("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);        }        int k1=0;        for (int i=0;i<n;i++){            for (int j=0;j<n;j++){                sum1[k1++]=a[i]+b[j];            }        }        int k2=0;        for (int i=0;i<n;i++){            for (int j=0;j<n;j++){                sum2[k2++]=c[i]+d[j];            }        }        sort (sum1,sum1+k1);        int count=0;        for (int i=0;i<k2;i++){                count+=upper_bound(sum1,sum1+k1,-sum2[i])-lower_bound(sum1,sum1+k1,-sum2[i]);//点睛之笔         }        printf ("%d\n",count);    }               return 0;}

其中有一个点睛之笔
count+=upper_bound(sum1,sum1+k1,-sum2[i])-lower_bound(sum1,sum1+k1,-sum2[i]);//点睛之笔
这是借鉴别人的成果，通过运用这个，可以减少很多不必要的操作。