4 Values whose Sum is 0
- Time Limit:
- 15000ms
- Memory limit:
- 228000kB
- 题目描述
- 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 .
- 输入
- 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 .
- 输出
- For each input file, your program has to write the number quadruplets whose sum is zero.
- 样例输入
-
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
- 样例输出
-
5
- 提示
- 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).
解答：#include <iostream>
#include <algorithm>
#define N 4000
using namespace std;
int a[N],b[N],c[N],d[N],ab[N*N],cd[N*N];
int main()
{
int i,j,n,pab,pcd;
while(cin>>n)
{
pab=pcd=0;
for(i=0;i<n;i++)
{
//cin>>a[i]>>b[i]>>c[i]>>d[i];
scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
ab[pab++]=a[i]+b[j];
cd[pcd++]=-(c[i]+d[j]);
}
sort(cd,cd+n*n);//是数组的起始地址与最后地址
int k,ans=0;
for(i=0;i<pab;i++)
{
int l=0,r=pcd-1,mid;
while(l<=r)
{
mid=(l+r)/2;
if(ab[i]==cd[mid])
{
ans++;//下一句怎么写
for(k=mid+1;k<pcd;k++)
if(ab[i]==cd[k]) ans++;
else break;
for(k=mid-1;k>=0;k--)
if(ab[i]==cd[k]) ans++;
else break; //因为已经排好序了，所以这里只要找最靠近Mid的数，不用continue,否则超时
break;
}
else if(ab[i]<cd[mid]) r=mid-1;
else l=mid+1;
}
}
cout<<ans<<endl;
}
// system("pause");
return 0;
}
