题面
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： 给一个n*4的矩阵，输入n*4个数，在每一列找一个数，使得四个数的和为0； 分析： 先分别求出a和b，c和d两列任意两个数的和存放到相应的数组，将cd的和进行排序后，再用二分法进行查找；二分查找的时候注意，倘若中间的数据符合条件的话要再往两边进行查找，因为不能排除有多个数字相等的情况 注意： 求第二组数据的时候，根据提交的结果是不需要初始化total的； 题解：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=4005;
int a[N],b[N],c[N],d[N];
int ab[N*N],cd[N*N];
int main()
{
int n,total=0,i,j;
while (cin>>n)
{
for (i=0;i<n;i++)
cin>>a[i]>>b[i]>>c[i]>>d[i];
int num1=0,num2=0;
for (i=0;i<n;i++)
for (j=0;j<n;j++)
{
ab[num1++]=a[i]+b[j];
cd[num2++]=-(c[i]+d[j]);
}
sort (cd,cd+num2);
for (i=0;i<num1;i++)
{
int mid,up=num2-1,low=0;
while (low<=up)
{
mid=low+(up-low)/2;
if (ab[i]==cd[mid])
{
total++;
for (j=mid+1;j<=up;j++)
{
if (ab[i]==cd[j])
total++;
else
break;
}
for (j=mid-1;j>=low;j--)
{
if (ab[i]==cd[j])
total++;
else
break;
}
break;
}
else
{
if (ab[i]>cd[mid])
low=mid+1;
else
up=mid-1;
}
}
}
cout << total << endl;
}
return 0;
}
